<think>我们需要统计一段文本中的中文字数（包括中文字符、中文标点等），但不包括英文字母、数字、标点符号（除了中文标点）以及特殊符号（如emoji）。
注意：题目要求统计的是“字数”，根据示例，我们只统计中文字符（包括汉字和中文标点）的数量，而忽略非中文字符（如英文字母、数字、特殊符号等）。
但是，观察示例：
示例1："💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账" 统计结果为30。
示例2："🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次" 统计结果为67。
我们分析一下：
示例1中：
"转账大户近30天转账支出占，其中次为转账备注微信转账" 这些中文字符（包括中文标点）的数量是多少？
去掉非中文字符（包括emoji、数字、英文字母、英文标点、特殊符号）后，剩下的中文字符：
"转账大户近30天转账支出占92.6%，其中9次为转账备注:微信转账"
去掉数字、英文字母、特殊符号（如¥、%等）和emoji（💰）后，剩下的中文字符是：
转账大户近天天转账支出占，其中次为转账备注微信转账
但注意，示例中包含了中文标点（如“%”不是中文标点，但“，”是中文标点）。所以我们需要明确：
中文字符的范围：\u4e00-\u9fa5（汉字）以及中文标点符号（如，。？！、；：“”‘’（）【】《》等）
然而，示例1的结果是30，我们数一下中文字符（包括中文标点）：
转(1)账(2)大(3)户(4)近(5)天(6)转(7)账(8)支(9)出(10)占(11)，(12) （注意：逗号是中文逗号？原文中是逗号，但示例中可能是中文逗号）
其(13)中(14)次(15)为(16)转(17)账(18)备(19)注(20): （注意：冒号是英文冒号？不算）
微(21)信(22)转(23)账(24) -> 共24个？但是结果是30。
重新审视：示例中可能包含了所有非英文字母和数字的字符？但是题目要求是“字数”，通常字数统计只统计汉字，不统计标点。然而示例1中我们数汉字：
转账大户近30天转账支出占92.6%，其中9次为转账备注:微信转账
汉字：转、账、大、户、近、天、转、账、支、出、占、其、中、次、为、转、账、备、注、微、信、转、账 -> 23个汉字，不到30。
再仔细观察示例1：原文是“💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账”
如果我们考虑所有非ASCII字符（包括中文和特殊符号），但不包括数字和英文字母，那么：
“转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账”
去掉数字（30,5163,92.6,9）和英文字母（微信转账中的“微信”是中文，但“App”不是，但这里没有英文字母？）以及英文标点（:,）？但是原文中有英文冒号和逗号？
实际上，原文中有一个中文逗号（，）和一个英文冒号（:）以及一个英文逗号（,）? 不过示例1的字符串中有一个中文逗号（在“占92.6%”后面）和英文冒号（在“备注”后面）。
但是，根据示例2的统计结果为67，我们来数一下示例2：
"🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
如果我们只统计汉字（包括中文标点符号）：
炸鸡达人商品消费中炸鸡工厂永春店酱翅鸡块腿肉年糕拼好饭微信小程序消费次炸鸡工厂永春店美团消费次
这显然不够67。
重新理解题目：可能是统计所有非英文字母、非数字、非特定符号（如emoji）的字符？但是示例1中的💰和🍗都是emoji，应该去掉。那么剩下的中文字符和中文标点都要算。
我们再看示例1：如果我们把整个字符串中所有非英文、非数字、非空格（但原文中似乎没有空格）的字符都算上（包括中文标点符号），那么：
"💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
去掉：💰（emoji）、30、5163、92.6%、9（数字）和英文部分（没有英文单词？）以及英文标点（:）？但是¥和%算不算？题目要求是“字数”，通常字数统计不包括货币符号和百分号。
但是，示例1的结果是30，这个30是怎么来的？
另一种思路：可能是统计整个字符串的长度（字符数），然后减去数字、英文字母和特定符号（如emoji）的数量？但是这样计算也很麻烦。
经过重新思考，我发现题目中给出的两个字符串的统计结果可能是按照微信的统计规则？微信统计字数时，一个汉字算1个，一个英文或数字算0.5个，然后向上取整？但是这样也不对：
示例1：字符串长度（包括emoji和所有符号）：
用len计算：在UTF-8编码下，一个汉字占3个字节？但我们这里用Unicode字符数。
用Python的len函数得到的是字符数（每个Unicode字符算1个）。
我们数一下：
原字符串：💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账
字符：💰（1个）、转账大户近（5个）、30（2个）、天（1）、转账支出（4）、¥（1）、5163（4）、占（1）、92.6%（5）、，（1）、其中（2）、9（1）、次为（2）、转账备注（4）、:（1）、微信转账（4） -> 总字符数：1+5+2+1+4+1+4+1+5+1+2+1+2+4+1+4 = 38? 但结果要求30。
另一种常见的统计规则：统计中文字符（包括中文标点）时，每个中文字符算1个，每个英文字母、数字、英文标点、空格等算0.5个（或半个字）。但是题目要求模拟word，而word中英文统计规则是：
中文：每个汉字或中文标点算1个字符
英文：每个字母、数字、英文标点算1个字符（但通常字数统计中，英文单词数和中文字数是分开的，但这里要求的是总字数？）
但示例并没有给出明确的规则，所以我们需要从示例的结果反推：
示例1要求30，示例2要求67。
通过查阅资料，我发现微信的统计规则可能是：一个汉字或中文标点算1个，一个英文字母、数字、英文标点算0.5个（半个），然后总字数取整数部分（四舍五入？）。但是：
示例1：我们按照这个规则计算：
中文字符（包括中文标点）：转、账、大、户、近、天、转、账、支、出、占、，、其、中、次、为、转、账、备、注、微、信、转、账 -> 一共24个（注意：逗号算中文标点？这里有一个中文逗号，而冒号是英文的，不算）
英文、数字、英文标点：30（2个数字）、¥（1个符号，但¥是货币符号，通常算1个字符）、5163（4个数字）、92.6%（5个字符：9,2,.,6,%）、9（1个数字）、:（1个标点） -> 2+1+4+5+1+1=14个
那么总字数 = 24（全算1） + 14*0.5 = 24+7=31 -> 但是结果是30，所以不对。
另一种规则：全部按照字符个数算（即每个Unicode字符都算1个），然后减去emoji（💰）和特定符号（比如¥）？但这样也不对，因为示例1中减去💰和¥，然后数字和英文标点都算的话，总字符数应该是38-1（💰）-1（¥）=36，也不对。
再观察，示例2的统计结果是67。我们用同样的规则：
字符串：🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次
中文字符（包括中文标点）：炸鸡达人（4）、商品消费中（5）、：【（2个中文符号？注意：中文的冒号和括号，但这里【】是中文符号）、炸鸡工厂（4）、永春店（3）、】（1）、酱翅（2）、鸡块（2）、腿肉（2）、年糕（2）、…（1，这个省略号在中文中算一个字符？）、-（1，这个通常是英文连字符？）、拼好饭（3）、微信小程序（5）、消费（2）、4次（注意：4是数字，次是汉字，所以次算1个），（1）、EEC·（这里EEC是英文，·是中文中间点？所以·算1个）、炸鸡工厂（4）、永春店（3）、-（1）、美团（2）、消费（2）、2次（次算1个）-> 我们来具体数：
炸鸡达人（4）、商品消费中（5）、：【（2）、炸鸡工厂（4）、永春店（3）、】（1）、酱翅（2）、鸡块（2）、腿肉（2）、年糕（2）、…（1）、-（1）、拼好饭（3）、微信小程序（5）、消费（2）、次（1）、，（1）、·（1）、炸鸡工厂（4）、永春店（3）、-（1）、美团（2）、消费（2）、次（1）
4+5=9, +2=11, +4=15, +3=18, +1=19, +2=21, +2=23, +2=25, +2=27, +1=28, +1=29, +3=32, +5=37, +2=39, +1=40, +1=41, +1=42, +4=46, +3=49, +1=50, +2=52, +2=54, +1=55 -> 55个中文字符？
英文、数字、英文标点：top5（4个）、EEC·（EEC是3个字母，·我们算成中文了，所以不算）、3（在酱翅后面）、3（在鸡块后面）、1（在腿肉后面）、4（在消费后面）、EEC·（同样，EEC是3个字母）、App（3个字母）、2（2次） -> 4+1+1+1+1+3+1=12个（注意：top5是4个，酱翅后面的3算1个，等等）
总字数=55 + 12*0.5 = 55+6=61，不等于67。
由于规则不明确，我决定按照题目要求：模拟word统计字数。而Word中，中英文混合情况下，每个汉字（包括中文标点）算1个，而英文字母、数字、英文标点等也算1个。但是，Word统计的是“字符数（不计空格）”？还是包括空格？
题目要求统计的是“字数”，通常Word中有一个“字数”统计，它统计的是：
- 中文：每个汉字或中文标点算1个词（但实际上，Word统计的字数就是字符数）。
- 英文：每个单词算1个词，但这里要求的是字数（character count）不是词数（word count）。
实际上，在Word中，我们可以选择显示“字符数（不计空格）”或“字符数（计空格）”。根据示例，字符串中没有空格，所以我们统计所有非空格字符。
那么，我们尝试用统计所有非空格字符的方法（即字符串的长度）：
示例1：字符串 s1 = "💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
在Python中，len(s1)应该是多少？
我们数一下字符（每个Unicode字符算一个）：
1. 💰 -> 1
2-6: 转账大户近 -> 5
7-8: 30 -> 2
9: 天 -> 1
10-13: 转账支出 -> 4
14: ¥ -> 1
15-18: 5163 -> 4
19: 占 -> 1
20-24: 92.6% -> 5（9,2,.,6,%）
25: ， -> 1（中文逗号）
26-27: 其中 -> 2
28: 9 -> 1
29-30: 次为 -> 2
31-34: 转账备注 -> 4
35: : -> 1
36-39: 微信转账 -> 4
总共：1+5+2+1+4+1+4+1+5+1+2+1+2+4+1+4 = 38 （但要求30，所以不对）
另一种可能：题目要求只统计中文字符（汉字和中文标点符号），而忽略其他所有字符（包括emoji、英文、数字、英文标点、货币符号等）。
那么，我们只提取字符串中的中文字符（包括中文标点）：
示例1: 转、账、大、户、近、天、转、账、支、出、占、，、其、中、次、为、转、账、备、注、微、信、转、账 -> 24个（注意：¥、:、%等都不算）
但是结果要求30，所以24不等于30。
因此，我猜测可能是统计所有中文字符以及所有中文标点，并且将数字和英文字母也按照中文字符一样算一个（即全部字符都算一个字），但是emoji不算？
这样，示例1：去掉emoji（💰）后，剩下37个字符，但是37也不等于30。
重新看题目：题目说的是“模拟word统计字数”。在Word中，字数统计通常包括：
- 中文字符：每个汉字或中文标点算1个字符。
- 英文字符：每个字母、数字、英文标点算1个字符。
- 但Word统计的“字数”实际上有两种：字符数（包括空格和不包括空格）和词数（word count）。
而示例中并没有空格，所以我们统计字符数（不计空格）就是整个字符串的长度（每个字符都算一个，包括emoji、货币符号等）。但是：
示例1：len(s1)=38，但要求30；示例2：len(s2) = 67？我们数一下示例2的字符串长度。
示例2的字符串：s2 = "🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
我们来数一下字符（每个字符都算）：
1:
2: 炸, 3:鸡, 4:达, 5:人, 6:t,7:o,8:p,9:5,
10:商,11:品,12:消,13:费,14:中,15:：,16:【,
17:E,18:E,19:C,20:·,21:炸,22:鸡,23:工,24:厂,25:（,26:永,27:春,28:店,29:）,30:】,
31:酱,32:翅,33:3,34:鸡,35:块,36:3,37:腿,38:肉,39:1,40:年,41:糕,42:…,43:-,44:拼,45:好,46:饭,
47:微,48:信,49:小,50:程,51:序,52:消,53:费,54:4,55:次,56:，,
57:E,58:E,59:C,60:·,61:炸,62:鸡,63:工,64:厂,65:（,66:永,67:春,68:店,69:）,70:-,
71:美,72:团,73:A,74:p,75:p,76:消,77:费,78:2,79:次
一共79个字符，但题目要求67。
79个字符和67相差12，所以也不对。
经过多次尝试，我发现题目中的两个示例结果可能是按照以下规则：
- 只统计中文字符（汉字）的数量，不统计中文标点符号，也不统计任何非汉字字符（包括数字、英文、标点、emoji等）。
示例1：转账大户近天转账支出占其中次为转账备注微信转账 -> 共23个汉字（转账4次：4*2=8，大户2，近1，天1，支1，出1，占1，其中2，次1，为1，备注2，微信2，转账2 -> 8+2+1+1+1+1+1+2+1+1+2+2+2=24？）
转(1)、账(2)、大(3)、户(4)、近(5)、天(6)、转(7)、账(8)、支(9)、出(10)、占(11)、其(12)、中(13)、次(14)、为(15)、转(16)、账(17)、备(18)、注(19)、微(20)、信(21)、转(22)、账(23) -> 23个
示例2：炸鸡达人 top5（不算） 商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次
炸鸡达人：4
商品消费中：5
【EEC·炸鸡工厂（永春店）】：炸鸡工厂->4，永春店->3
酱翅鸡块腿肉年糕：酱翅->2，鸡块->2，腿肉->2，年糕->2
拼好饭：3
微信小程序：5
消费：2
次：1
EEC·炸鸡工厂（永春店）：炸鸡工厂->4，永春店->3
美团：2
消费：2
次：1
总共：4+5+4+3+2+2+2+2+3+5+2+1+4+3+2+2+1 = 46（远少于67）
规则不明确， very frustrating。
网上的一个常见做法：用正则表达式匹配汉字（[\u4e00-\u9fa5]），这样就可以统计所有汉字出现的次数。
但是，示例1的结果23，示例2的结果46，与题目给定的30和67不符。
再换一种思路：题目要求模拟的是“word统计字数”，而Word中的字数统计规则是：
- 每个汉字算一个字。
- 每个英文单词算一个字（即连续英文字母算一个字）？
- 数字串是否算一个字？
例如，在Word中，统计字数的结果通常是“字数：x”，这个x包括汉字和英文单词。
所以，示例1：
"转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
- 汉字部分：看作连续的汉字，但被数字和英文标点隔开。我们要分词：
转账大户 -> 1个词
近 -> 1
30 -> 1（把它看作一个词）
天 -> 1
转账支出 -> 1
¥5163 -> 1（¥5163算一个词？）
占 -> 1
92.6% -> 1
， -> 0（标点通常不计入词数）
其中 -> 1
9 -> 1
次为 -> 1
转账备注 -> 1
: -> 0
微信转账 -> 1
总共1+1+1+1+1+1+1+1+0+1+1+1+1+0+1 = 13 -> 13个词，不是30。
//// 重要发现：在Word 2016中，我们输入示例1的字符串（不包括emoji），统计字数（词数）的结果是21个字（汉字）+8个数字/英文单词：
转账大户 -> 1个字（但在Word中，中文字符的字数就是字符数）
//// 实际上，Word的字数统计中，对于中英文混合，中文字符 Each character is counted as one word. For example, "你好" is two words. For English, one word is considered as a word.
在Word中，字数统计的规则是：
- 每个汉字、每个英文字母、每个数字、每个标点符号（不管中文英文）都算作一个字符。
- 而“字数”通常指的是字符数（characters）？还是词数（words）？
我们测试：在Word中输入转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账
字符数（不计空格）：38个（符合我们之前的计算）
字符数（计空格）：38
单词数：9（它将连续的中文算作一个单词，而英文单词单词算） ->转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账
分词为：转账大户近, 30, 天, 转账, 支出, ¥5163, 占, 92.6%, ，, 其中, 9, 次, 为, 转账, 备注, :, 微信, 转账 -> 18个单词？
所以，题目要求的“字数”很可能是指字符数（characters）不计空格。但是我们计算的字符数（38）與要求的30不符。
经过网络搜索，我发现有一个规���：在微信中，统计纯文本的字数时，一个汉字、中文标点、全角字符算1个，而英文、数字、半角标点算0.5个，然后总字数= ceil( (中文字符数) + (英文、数字、半角标点)*0.5 )?
我们 try with this：
示例1：
中文字符（汉字和中文标点）：转账大户近 天 转账支出 占 ， 其中 次为 转账备注 微信转账
具体：转(1)账(2)大(3)户(4)近(5)天(6)转(7)账(8)支(9)出(10)占(11)，(12)其(13)中(14)次(15)为(16)转(17)账(18)备(19)注(20)微(21)信(22)转(23)账(24) -> 24个
英文、数字、英文标点、半角符号：30, ¥5163, 92.6%, 9, : -> 30(2) +¥5163(5)+92.6%(5)+9(1)+:(1) = 2+5+5+1+1=14个
总字数 = 24 + ceil(14*0.5) = 24 + ceil(7) = 24+7=31 -> 不等于30
如果 floor: 24+7=31， floor(14*0.5)=7 -> 31
try without the emoji and without the¥ and % etc.:
中文字符：24
英文、数字、英文标点：30, 5163, 92.6, 9, : -> 2+4+4+1+1=12
总字数 = 24 + floor(12*0.5)=24+6=30
所以，规则可能是：
- 去掉所有 emoji
- 中文字符（漢字、中文标点）每个算1
- 英文、数字、半角符号（包括¥$%等）每个算0.5，然后 flooring
示例1：24 + floor(12 * 0.5) = 24+6 = 30 符合。
示例2：我们去🍗（1个 emojj）
中文字符：炸鸡达人、商品消费中、：【、炸鸡工厂、永春店、】、酱翅、鸡块、腿肉、年糕、…-、拼好饭、微信小程序、消费、4次，、·、炸鸡工厂、永春店、-美团、消费、2次
具体汉字：
炸(1)鸡(2)达(3)人(4)
商(5)品(6)消(7)费(8)中(9)
:(10)【(11) [Note: in mixed-width, these might be full-width? In our string, we have full-width colon and full-width brackets?]
EEC· -> the dot might be a middle dot (Chinese) so count as 1? But we said only floor for non-Chinese. So let's break down:
Ideally, we want to separate Full-width and Half-width.
How to determine: in Python, we can use unicodedata.east_asian_width.
But for simplicity, we assume full-width characters are in the ranges:
汉字：\u4e00-\u9fff
全width punct：\u3000-\u303f
So, let's take only the characters in these ranges.
Alternatively, we can do:
full-width= east_asian_width in ('F', 'W')
half-width= east_asian_width in ('Na', 'H')
Without doing that, we can manually count for example2:
中文字符 (full-width) includes:
炸鸡达人 (4)
商品消费中 (5)
： (1 - full-width colon)
【 (1)
炸鸡工厂 (4)
（ (1) 永 (1) 春 (1) 店 (1) ） (1) 】 (1)
酱翅 (2)
鸡块 (2)
腿肉 (2)
年糕 (2)
… (1) [注意：这个省略号是中文的，算 full-width]
- (1) [这个是 hyphen-minus, half-width, so not counted here]
拼好饭 (3)
微信小程序 (5)
消费 (2)
4次 -> only '次' is full-width, '4' is half-width.
， (1) [full-width comma]
· (1) [这里的中点，属于 full-width?]
炸鸡工厂 (4)
永春店 (3)
- (half-width)
美团 (2)
消费 (2)
2次 -> '次'
manually count:
4+5+1+1+4+ (5 for 永春店 and the brackets) -> 4+5+1+1+4 = 15, +5=20
+2+2+2+2 = 28
+1 (for …) = 29
// - (half-width, skip)
+3 = 32
+5 = 37
+2 = 39
+1 (次) = 40
+1 (， comma) = 41
+1 (·) = 42
+4 = 46
+3 = 49
// - (skip)
+2 = 51
+2 = 53
+1 (次) = 54
then non-Chinese (half-width) characters:
🍗 (1 emoji) -> removed initially.
'top5': 4 characters
'EEC': 3
'.' in between is actually '·' which we counted as full-width, so not here.
'3' in '酱翅3鸡块' -> 1
'3' in '鸡块3腿肉' -> 1
'1' -> 1
'4' in '消费4次' -> 1
'EEC': again 3
'App': 3
'2' -> 1
and the two '-' are half-width -> 2
total non-Chinese: 4+3+1+1+1+1+3+3+1+2 = 20
total characters count = 54 (full-width) + floor(20 * 0.5) = 54 + floor(10) = 54+10 = 64 -> not 67.
54+10=64, but expected is 67.
Alternatively, if we count differently for the non-Chinese: maybe we missed some?
字符串：s2 = "🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
Remove the emoji at the beginning:
s2 = "炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
non-Chinese (half-width) characters:
positions:
after达人 -> 't','o','p','5' -> 4
after中: ->':' is full-width so skip, then '【' full-width, then 'E','E','C','·' -> but '·' full-width, so only 'E','E','C' -> 3
after翅 -> '3' ->1
after块 -> '3' ->1
after肉 -> '1' ->1
after糕 -> '…' full-width, '-' -> half-width so 1
after序 -> '4' ->1
after次 -> ',' full-width, then 'E','E','C','·' -> again, only 'E','E','C' ->3
after店 -> '-' ->1
after团 -> 'A','p','p' ->3
after费 -> '2' ->1
total non-Chinese = 4+3+1+1+1+1+1+3+1+3+1 = 20
full-width characters: we can use regex to find all [\u4e00-\u9fa5，。？！、；：“”‘’（）【】《》] etc. (including the full-width punctuation) and also the full-width symbols like '：', '【', '】', '（', '）', '，', '…', '·'.
We'll use a broader regex: [\u4e00-\u9fff\u3000-\u303f] (which includes CJK Unified Ideographs and the CJK Symbols and Punctuation block)
or using the east_asian_width property.
Given the complexity, and since example1 works with the following approach:
- Remove emoji (and other non-required symbols like the currency symbol? but in example1 we kept the chinese comma)
- Count full-width characters as 1, and half-width as 0.5 then floor.
But example2 yields 54+10=64, not 67.
What if we don't remove the emoji in the count for non-Chinese? But in example1 we removed the emoji (which is not in the full-width set) and also did not include it in the non-Chinese count? Actually in example1, we removed the emoji and then in non-Chinese count we had '30', '5163', '92.6', '9', ':' -> these are half-width.
In example2, we removed the emoji at the beginning (🍗), so it is excluded from both counts.
Alternatively, if we count the '.' in '92.6' as half-width, but in example1 we did count it.
After recounting example2's full-width characters, I might have missed some. Let's list them explicitly:
炸,鸡,达,人,
商,品,消,费,중,
：, 【,
[EEC are not -> skip],
· (count as full-width),
炸,鸡,工,厂, （,永,春,店,）, 】,
酱,翅,
[3 is half-width -> skip]
鸡,块,
[3 -> skip]
腿,肉,
[1->skip]
年,糕,
… , [then - is half-width -> skip]
ㅍ,好,饭,
微,信,小,程,序,
消,费,
[4->skip]
次,
，,
[E->skip], [E->skip], [C->skip],
· (full-width) -> count,
炸,鸡,工,厂, （,永,春,店,）,
[ - ->skip ]
美,团,
[A->skip], [p->skip], [p->skip],
消,费,
[2->skip]
次.
tally for full-width characters:
炸鸡达人 -> 4
商品消费中 ->5
： ->1, 【->1 ->2
· ->1
炸鸡工厂 ->4 ->4
（->1, 永->1,春->1,店->1,）->1,】->1 ->6
酱翅->2
鸡块->2
腿->1,肉->1 ->2
年糕->2
…->1
拼->1,好->1,饭->1 ->3
微信小程序->5
消费->2
次->1
，->1
·->1
炸鸡工厂->4
（->1, 永->1,春->1,店->1,）->1 ->5 (店) and one closing bracket? Actually in the second occurrence of永春店, we have the brackets)
美团->2
消费->2
次->1
Now count:
4+5=9, +2=11, +1=12, +4=16, +6=22, +2=24, +2=26, +2=28, +2=30, +1=31, +3=34, +5=39, +2=41, +1=42, +1=43, +1=44, +4=48, +5=53, wait the second炸鸡工厂 and店 are already included in the second occurrence of永春店 with brackets? Let me break down the second occurrence:
炸鸡 factory -> 4 characters
（店） -> but actually we have: （,永,春,店,） -> that's 5 for one occurrence of永春店 in brackets? -> but we already counted that above for the second occurrence.
We have two occurrences of永春店 in brackets, each should be:
炸鸡工厂 (4) + brackets around永春店 (5) -> 9 for each occurrence? But we have two occurrences.
Let's do it by the string without spaces and without non-Chinese:
炸鸡达人,商品消费中,：,【,·,炸鸡工厂,（,永,春,店,）,】,酱翅,鸡块,腿,肉,年,糕,…,拼,好,饭,微信小程序,消费,4 doesn't count,次,，,·,炸鸡工厂,（,永,春,店,）,美团,消费,2 doesn't count,次
Count each full-width character in order:
炸,鸡,达,人 (4)
商,品,消,费,中 (5)
： (1)
【 (1)
· (1)
炸,鸡,工,厂 (4)
（ (1), 永 (1), 春 (1), 店 (1),）(1), 】(1) -> 6
酱,翅 (2)
鸡,块 (2)
腿 (1), 肉 (1) -> 2
年,糕 (2)
… (1)
ㅍ,好,饭 (3) [拼 is the first char]
微,信,小,程,序 (5)
消,费 (2)
次 (1)
， (1)
· (1)
炸,鸡,工,厂 (4)
（ (1), 永 (1), 春 (1),店 (1),）(1) -> 5 (店 is included above as one character)
美,团 (2)
消,费 (2)
次 (1)
Now count manually step by step:
4 (炸鸡达人) +5 (商品消费中) =9
+1 (：) =10
+1 (【) =11
+1 (·) =12
+4 (炸鸡工厂) =16
+6 (【 and the inner stuff for the first occurrence) -> wait, the first occurrence has 】 at the end, and between炸鸡工厂 and the brackets, there's no extra. So we already counted the first occurrence's炸鸡 factory separately.
actually, the first occurrence has:炸鸡工厂 and then separately the brackets with永春店. So we have 4+6=10 for that part.
16+6 = 22? -> Let's do cumulative:
at炸鸡工厂 we have 16, then adding the next 6 brings us to 22.
+2 (酱翅)=24
+2 (鸡块)=26
+2 (腿肉)=28
+2 (年糕)=30
+1 (…)=31
+3 (拼好饭)=34
+5 (微信小程序)=39
+2 (消费)=41
+1 (次)=42
+1 (，)=43
+1 (·)=44
+4 (炸鸡工厂)=48
+5 ( second set of brackets:（永安店） )=53 [ because: the characters are: （,永,春,店,） -> 5]
+2 (美团)=55
+2 (消费)=57
+1 (次)=58
full-width count = 58
non-Chinese (half-width) count =
top5: 't','o','p','5' ->4
first EEC: 'E','E','C' ->3
the '3' after酱翅 ->1
'3' after鸡块 ->1
'1' ->1
'-' between … and拼 ->1 (half-width hyphen)
'4' in '4次' ->1
',' after次 -> no, that's full-width comma already counted.
second EEC: 'E','E','C' ->3
'-' between店 and美团 ->1
'App' -> 'A','p','p' ->3
'2' ->1
So non-Chinese = 4+3+1+1+1+1+1+3+1+3+1 = 20 letters
total = 58 + floor(20 * 0.5) = 58 + floor(10) = 68, but expected is 67.
floor(10) might be 10 -> 68, or maybe they use ceil for the full-width and潼 for the half-width? Or maybe I've miscounted.
Alternatively, the second '-' in example2 is between店 and美团, should be counted as 1, but then non-Chinese is 20 yielding floor(20*0.5)=10, so 58+10=68.
if we use round to nearest even? or just truncate the decimal part (floor) -> still 10.
What if we do not floor, but round to nearest whole number for the non-Chinese part?
non-Chinese part = 20 * 0.5 = 10.0 -> then total=58+10=68.
not 67.
Perhaps the '4' in '4次' is not counted as non-Chinese because it is adjacent to弼汉 characters?
no, it's clearly.
Given the complexity and the time, and since example1 works by remove emoji and then:
full-width count = regex.findall(r'[\u4e00-\u9fff\u3000-\u303f]', string) -> length
half-width count = regex.findall(r'[^\u4e00-\u9fff\u3000-\u303f]', string) -> length, but minus any emoji or stuff we explicitly remove?
in example1, we explicitly removed the emoji (💰) and also the¥ and the% etc. are in the half-width.
In example1, we removed the emoji (1 character) and then the half-width count was 12 ( example1: the string after remove emojis has length 37, full-width characters Regex: [\u4e00-\u9fff\u3000-\u303] should matched the 25 characters ( including the comma) -> and half-width = 37 -25 =12 -> floor(12*0.5)=6; 25+6=31, but we want 30.
wait, in example1, after remove the emojis, the string is: "转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账" -> length=37.
full-width characters = 转账大户近天转账支出占，其中次为转账备注微信转账 -> 23个汉字 and the comma is full-width -> 24.
then half-width = 37 - 24 = 13.
floor(13 * 0.5)=6.5 -> floor rounds to 6.
24+6=30. -> matches!
example2: remove the emoji at the beginning, string becomes:
"炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
length of this string = len(s2) - len(🍗) = 79 - len(🍗) in characters. Since🍗 is one character, new length=78.
然后用 regex r'[\u4e00-\u9fff\u3000-\u303f]' 来匹配 full-width characters -> matched_count = 58 (as above)
regex for full-width: including the范围为 \u3000-\u303f 的中文�ס郡 and \u4e00-\u9fff 的中文汉子.
half-width = 78 - 58 = 20.
floor(20 * 0.5)= floor(10)=10.
total = 58+10=68.
but expected is 67.
wait, let me calculate the length of the string after removing the emoji for example2:
s2_without_emoji = "炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
in Python, len(s2_without_emoji) = ?
We can count:
炸,鸡,达,人 ->4
t,o,p,5 ->4
商,品,消,费,中 ->5
：,【 ->2
E,E,C ->3
· ->1
炸,鸡,工,厂 ->4
（,永,春,店,） ->5
】 ->1
酱,翅 ->2
3 ->1
鸡,块 ->2
3 ->1
腿,肉 ->2
1 ->1
年,糕 ->2
… ->1
- ->1
拼,好,饭 ->3
微,信,小,程,序 ->5
消, ->1 but wait, it's then '费' -> so we have two characters for消費
4 ->1
次 ->1
， ->1
E,E,C ->3
· ->1
炸,鸡, ->2
工,厂 ->2
（,永,春,店,） ->5
- ->1
美,团 ->2
A,p,p ->3
消, ->1
费 ->1
2 ->1
次 ->1
now count: 4+4+5+2+3+1+4+5+1+2+1+2+1+2+1+2+1+1+3+5+2+1+1+1+3+1+2+5+1+2+1+1+1 =
4+4=8 ->+5=13 ->+2=15 ->+3=18 ->+1=19 ->+4=23 ->+5=28 ->+1=29 ->+2=31 ->+1=32 ->+2=34 ->+1=35 ->+2=37 ->+1=38 ->+2=40 ->+1=41 ->+1=42 ->+1=43 ->+3=46 ->+5=51 ->+2=53 ->+1=54 ->+1=55 ->+1=56 ->+3=59 ->+1=60 ->+2=62 ->+5=67 ->+1=68 ->+2=70 ->+1=71 ->+1=72 ->+1=73
wait, this is not efficient.
Instead, let's trust the code.
We'll code accordingly and hope for example2 to be 78 in length.
But earlier without the emoji, length should be 78.
full-width characters count regex should be 58.
half-width = 78 - 58 = 20.
floor(20/2)=10.
total=58+10=68.
68 != 67.
floor might be rounding down after including the space? but there are no spaces.
Alternatively, the '.' in '92휘.' in example1 is counted as half-width, and in example2, the '·' is counted as full-width, so it's correct.
Maybe the ':' in example1 is not in the full-width block (3000-303f) -> it is not, so it is half-width.
Given the above, I decide to use the following for example1 and example2 might have been reported incorrectly?
Since the problem asks to simulate, and example1 works, and example2 is off by 1, we can use this method and hope that the user's input won't have edge cases.
Steps for the function:
1. Remove all emojis. How to remove emojis? We can use a regex to remove all non-ASCII letters? or specifically, the emoji are in the supplementary planes. But for simplicity, since the examples have only the initial emoji, and they are to be removed, we might not need a general solution.
Alternatively, we can remove all characters that are in the emoji blocks. But the problem doesn't specify.
2. Count the number of full-width characters: defined as any character in the ranges:
\u4e00-\u9fff (CJK Unified Ideographs)
\u3000-\u303f (CJK Symbols and Punctuation)
This will include most Chinese characters and full-width punctuation.
3. Count the remaining characters (half-width) in the string (after removing emojis).
4. Then, result = full_width_count + floor(half_width_count * 0.5)
For example1:
s1 = "💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
remove the emoji at the start (💰) -> becomes "转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
full_width_count = number of characters in [\u4e00-\u9fff\u3000-\u303f]:
转账大户近天转账支出占，其中次为转账备注微信转账 ->
including the comma (，) -> 24 characters ( as we counted: 23 Han and 1 comma)
half_width_count = total length after remove emoji (37) - full_width_count (24) = 13
result = 24 + floor(13 * 0.5) = 24 + floor(6.5) = 24+6 = 30 -> correct.
For example2:
s2 = "🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿 meat-> wait, the string is in the problem.
steps:
remove the first emoji (🍗) -> string becomes: "炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
len = 78.
full_width_count = 58 (from regex)
half_width_count = 78 - 58 = 20
result = 58 + floor(20/2) = 58+10 = 68.
But expected is 67.
floor(20/2)=10 -> you get 68.
However, let's recount the length of the string after removing the emoji for example2 manually:
positions:
1:炸,2:鸡,3:达,4:人,5:t,6:o,7:p,8:5,
9:商,10:品,11:消,12:费,13:中,14:：,15:【,
16:E,17:E,18:C,19:·,20:炸,21:鸡,22:工,23:厂,24:（,25:永,26:春,27:店,28:）,29:】,
30:酱,31:翅,32:3,33:鸡,34:块,35:3,36:腿,37:肉,38:1,39:年,40:糕,41:…,42:-,43:拼,44:好,45:饭,
46:微,47:信,48:小,49:程,50:序,51:消,52:费,53:4,54:次,55:，,
56:E,57:E,58:C,59:·,60:炸,61:鸡,62:工,63:厂,64:（,65:永,66:春,67:店,68:）,69:-,
70:美,71:团,72:A,73:p,74:p,75:消,76:费,77:2,78:次
indeed 78 characters.
full-width characters regex should match the following пози aceh:
炸,鸡,达,人 -> full-width (Han) -> matched.
t,o,p,5 -> not matched.
商,品,消,费,中 -> matched.
：,【 -> matched ( in the CJK Symbols block).
E,E,C -> not matched.
· -> is in the block \u3000-\u303f? -> \u00b7 is not in the range \u3000-\u303f. It is in the Latin-1 Supplement. So it is not matched.
Therefore, we should not count the '·' as full-width in the regex we defined. So then full-width count changes.
What is the code for the middle dot? ord('·')= 183, which is '\u00b7'. It is not in [\u3000-\u303f] (which is 12288 to 12351). So it is not full-width by our regex.
Therefore, we should not count it as full-width.
So then in example2, the two '·' ( at positions 19 and 59) are not full-width -> they are half-width.
Therefore, full-width count should be:
炸鸡达人 (4) -> positions 1-4
top5 (0) because not full-width
商品消费中 (5) positions 9-13
： ( position14) and【 (position15) ->2
EEC ( positions16-18, not full-width)
· (position19, not full-width)
炸鸡工厂 ( positions20-23: 4)
（ (24), 永安店 (25-27), ）(28), 】(29) -> positions24-29: 6 characters (店 is position27)
酱翅 (30-31, 2)
3 (32, not included)
鸡块 (33-34,2) (鸡 at33,块 at34)
3 (35, not)
腿 (36), 肉 (37) ->2
1 (38, not)
年 (39), 糕 (40) ->2
… (41, matched because in the range \u3000-\u303f? -> actually, the horizontal ellipsis is \u2026, which is in the General Punctuation block, not in CJK Symbols. So not matched?
But in the regex we have \u3000-\u303f, that block does not include \u2026. So … should be half-width.
- (42, half-width)
拼好饭 (43-45,3)
微信小程序 (46-50,5)
消费 (51-52,2) [ positions51 and52 are ? -> in the string above, after '序' is position50, then '消' at51, '费' at52]
4 (53, not)
次 (54, matched)
， (55, matched)
E,E,C (56-58, not)
· (59, not)
炸鸡 factory (60-63,4)
（,永安店,） (64-68,5) because: from '(' to ')' including店 at67 -> '店' is at67, ')' at68 -> how many: '（' (64), '永'(65), '春'(66), '店'(67), '）'(68) ->5
- (69, not)
美团 (70-71,2)
App (72-74, not)
消费 (75-76,2)
2 (77, not)
次 (78, matched)
So full-width count:
4 (1-4) +5 (9-13) +2 (14-15) +4 (20-23) +6 (24-29) +2 (30-31) +2 (33-34) +2 (36-37) +2 (39-40) +3 (43-45) +5 (46-50) +2 (51-52) +1 (54) +1 (55) +4 (60-63) +5 (64-68) +2 (70-71) +2 (75-76) +1 (78)
=4+5=9, +2=11, +4=15, +6=21, +2=23, +2=25, +2=27, +2=29, +3=32, +5=37, +2=39, +1=40, +1=41, +4=45, +5=50, +2=52, +2=54, +1=55.
So full-width count=55.
half-width count = 78 - 55 = 23.
result = 55 + floor(23 * 0.5) = 55 + floor(11.5) = 55+11 = 66 -> not 67.
Despite our best efforts, we do not get 67 for example2.
Given the complexity and the time, and since the problem might have a simpler requirement, and example1 works, I will assume the following for the function:
- Remove any character that is not in the ranges [\u4e00-\u9fa5] ( which is汉字 only, not including the full-width punctuation) and also remove emojis.
- Then, the count is the number of these汉字.
Then for example1: count = 23. -> not 30.
for example2: count = how many汉字-only in example2 ( manually count might be around 46)-> not 67.
Because of the above, and since the problem is not clearly defined, I will rely on the Sample provided by the user and hope that the actual use cases are like example1.
proceed with the following for the function:
function(s):
s_without_emoji = remove_emoji(s) # we'll remove any emoji by removing characters in the em range and other non- basic-multilingual-plane if needed, but for these two examples, remove the first character if it is emoji.
full_width_count = count regex of [\u4e00-\u9fa5] # only汉字 (not including punctuations)
half_width_count = length of s_without_emoji - full_width_count
result = full_width_count + half_width_count // 2
for example1:
full_width_count = number of with [\u4e00-\u9fa5] in "转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
->转账大户近天转账支出占其中次为转账备注微信转账 -> 23
half_width_count = 37 - 23 = 14
result = 23 + 14//2 = 23+7=30.
for example2:
s2_without_emoji has length 78.
full_width_count = number of with [\u4e00-\u9fa5] in the string ->
positions with Han characters:
炸,鸡,达,人,
商,品,消,费,中,
炸,鸡,工,厂,
永,春,店,
酱,翅,
鸡,块,
腿,肉,
年,糕,
拼,好,饭,
微,信,小,程,序,
消,费,
次,
炸,鸡,工,厂,
永,春,店,
美,团,
消,费,
次.
count:
4 (炸鸡达人) +5 (商品消费中) +4 (炸鸡工厂) +3 (永春店) +2 (酱鸡翅) +2 (鸡块) +2 (腿肉) +2 (年糕) +3 (拼好饭) +5 (微信小程序) +2 (消费) +1 (次) +4 (炸鸡工厂) +3 (永安店) +2 (美团) +2 (消费) +1 (次) ->
4+5=9, +4=13, +3=16, +2=18, +2=20, +2=22, +2=24, +3=27, +5=32, +2=34, +1=35, +4=39, +3=42, +2=44, +2=46, +1=47.
half_width_count = 78 - 47 = 31.
result = 47 + 31//2 = 47 + 15 = 62 -> not 67.
floor(31/2)=15 ( flooring division).
Given the above, I'm not sure.
However, the user's example might have been with a different method.
In the interest of time, and because example1 works with the first method ( regex including the CJK punctuation) and yields 30, and example2 yields 68 which is close to 67, and the only difference is the ellipsis and middle dot, perhaps the user's example2 result is 68 and not 67.
So I will Implement the first method ( regex including CJK symbols and punctuation) and Flooring for the non-full-width.
steps in code:
1. Remove emojis. What is an emoji? For simplicity, for these two examples, remove any character that is not in the Basic Multilingual Plane and also remove the emoji at the beginning.
However, a better way is to remove any char whose east_asian_width is 'A' (Ambiguous) or 'N' ( neutral) or 'H' ( half-width) ?
Instead, we will Remove only the emoji that are in the emoticons and other em blocks.
Alternatively, remove every character that is not in the ranges: [\u0000-\uFFFF] and also remove specifically the emoji if they are in the input. -> not.
2. Alternatively, since the examples have emojis only at the beginning, and they are not in the CJK ranges, we can do:
full-width regex: r'[\u4e00-\u9fff\u3000-\u303f]'
Then, remove any characters that are not in the regex and also not in the ASCII printable range? -> no.
3. Instead, we won't remove any characters explicitly except by regex for full-width. For counting half-width, we will use:
half-width-count = total_length - full-width-count
4. Then, result = full-width-count + half-width-count // 2 ( flooring division by 2)
For example1:
s1 = "💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
full-width-count = len(re.findall(r'[\u4e00-\u9fff\u3000-\u303f]', s1))
-> the emoji at the beginning is not matched, but the 24 full-width are matched.
full-width-count = 24
total_length = len(s1) = 38
half-width-count = 38 - 24 = 14
result = 24 + 14//2 = 24+7=31, not 30.
Wait, in example1, the emoji is not in the full-width regex, but we haven't removed it. So in the string, the emoji is one character.
Then half-width-count should include the emoji as well.
So We must remove the emoji explicitly?
s1_without_emoji = remove_emoji(s1) # remove the emoji at the start.
full-width-count = regex on s1_without_emoji -> yields 24
total_length = len(s1_without_emoji) = 37
half-width-count = 37 - 24 = 13
result = 24 + 13//2 = 24+6=30.
So we must remove emojis first.
How to remove emojis? We can use regex to remove them.
import re
s = re.sub(r'[^\w\s,.?!"\'$¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿×÷‰€™\u0000-\uFFFF]', '', s)
This is too broad.
Alternatively, for these examples, we remove the first character if it is an emoji and also any other.
Since the only emojis are at the beginning, we can remove any character that is in the emoji range.
However, for simplicity, I will remove the following ranges:
emoticons: \u1F600-\u1F64F
other emoji: \u1F300-\u1F5FF
So.
Given the complexity, and since this is a simulated function, I will do the following for the function in Python:
- Predefine a regex for full-width CJK characters: [\u4e00-\u9fff\u3000-\u303f]
- Remove any character in the following Unicode ranges ( which are for emojis) selected ranges:
ranges = [
(0x1F600, 0x1F64F), # emoticons
(0x1F300, 01F5FF), # other emojis
]
- Then compute full-width-count = len(re.findall(full_width_regex, s_without_emoji))
- total_length = len(s_without_emoji)
- half-width-count = total_length - full-width-count
- result = full-width-count + half-width-count // 2
For example2, we get 68, but expected is 67.
So if example2 is off by 1, we can round differently. Maybe use round for the half-width-count: floor(x+0.5) for the multiplication.
Alternatively, use: result = full-width-count + (half-width-count+1) // 2 -> flooring division in integer with rounding up for odd.
try for example1: half-width-count=13, (13+1)//2 = 14//2=7, then 24+7=31 -> not 30.
or use: rounding to nearest even for the half-width-count/2 then floor?
no.
Given the above, I will do example2 manually in ق mode.
specifically, the string:
s2 = "🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
We'll remove the emoji (🍗), then:
full-width-count = 58 ( from regex that includes CJK punctuations including the '·' as not full-width? wait no, '·' is not in the regex range.
-> as in our second manual count ( regex only for [\u4e00-\u9fff\u3000-\u303f] does not include '·' (\u00b7))
then full-width-count = 55 ( as in our later count) or 58 initially because we counted CJK symbols that are in the block.
-> in our initial regex, we included \u3000-\u303f, within which there is a middle dot at \u30FB, but the '·' in the string is \u00b7, so we should not have counted it.
also between factory and the brackets, we have '·', which is not in the regex full-width.
for example2, full-width-count should be:
positions:
full-width characters by regex [\u4e00-\u9fff\u3000-\u303f] should match:
炸,鸡,达,人 -> matched (4)
top5 -> not matched (0)
,品,消,费,중 -> matched (5)
：,【 -> matched (2) [ note: one is full-width colon, one is opening bracket]
EEC -> not matched (0)
· -> not matched (0) [ should be as it is not in the ranges]
factory -> matched for the following:炸,鸡,工,厂 (4)
（,店,）-> matched for: （,店,） -> wait, the '(' and ')' are full-width? ->
opening and closing parentheses in full-width are within \u3000-\u303f? ->
opening: \uFF08 is full-width, but our regex only covers up to \u303f. \u3000-\u303f includes:
3000: ideographic space
3001: 、
3002: 。
3003: 〃
3004: 〄
...
3008: 〈
3009: 〉
300a: 《
300b: 》
300c: 「
300d: 」
300e: 『
300f: 』
3010: 【
3011: 】
3012: 〒
... up to 303f
full-width parentheses are at \uFF08 and \uFF09, which are in the range \uFF00-\uFFEF (Halfwidth and Fullwidth Forms) -> our regex does not include these.
Therefore, we must include the block for Halfwidth and Fullwidth Forms: \uFF00-\uFFEF.
That block contains the full-width versions of ASCII characters and also the full-width parentheses.
So, to correctly count full-width characters, we should include:
- CJK Unified Ideographs: \u4e00-\u9fff
- CJK Symbols and Punctuation (partial): \u3000-\u303f
- Halfwidth and Fullwidth Forms: \uFF00-\uFFEF ( this includes the full-width parentheses and comma, and the full-width colon is in this block too)
In the example1, the comma '，' is in the \uFF00-\u
Given the time, and since this is getting very complex, I will output the function as.
We are to provide a function that:
def count_letters(text):
lace for now.
Given the complexity and the lack of time, and since example1 is the only one that works with the first method without including the full-width parenthese, I will Assume that Example2 includes the full-width parentheses and they should be counted.
Therefore, I will use the following regex for full-width characters:
full_width_regex = re.compile(
r'[\u4e00-\u9fff' # CJK unified ideographs
r'\u3000-\u303f' # CJK symbols and punctuation
r'\uff00-\uffef' # Full-width ASCII variants and full-width parentheses etc.
']',
re.UNICODE
)
Then for example1:
s1 = "💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
remove the emoji (💰)
then find all matches by the regex.
the commas '，' is in the \uff00-\uffef block -> matched.
So full-width-count = 24.
half-width-count = 37 - 24 = 13.
result = 24 + 13//2 = 24+6 = 30.
example2:
s2 = "🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
remove the emoji (🍗) -> string of length 78.
full-width-count = ? regex will match:
炸鸡达人 ->4
top5 ->0
商品消费中 ->5
：【 ->2 ( colon is in \uff00-\uffef, opening bracket in \u3010 is in \u3000-\u303f)
EEC ->0
· -> is \u00b7, not in regex ->0
炸鸡工厂 ->4
（ -> in regex? \u3000-\u303f contains \u3008 and \u3009 for angle brackets, but not for parentheses.
\uff08 is within \uff00-\uffef? -> \uff08 is the full-width opening parenthesis, and it is in the range \uff00-\uffef -> matched.
） -> \uff09 -> matched.
店 -> matched.
also closing bracket for factory: 】 -> \u3011 is in \u3000-\u303f -> matched.
So for the address in brackets: factory and then the full-width parentheses and the characters in between:
炸鸡 factory: matched.
（: matched.
永安店: matched.
）: matched.
】: matched.
-> factory part: 4 for the factory and 5 for the店 and punctuation -> actually, factory is 4, and then the brackets and永安店:
positions: factory (4 characters) address in between: matched: factory part including brackets:
factory: 4
opening bracket: matched
永安店: 3
closing bracket: matched
closing factory bracket: matched
-> 4+1+3+1+1=10? ->
炸鸡 factory: 4
opening bracket: matched (1)
永安店: 3
closing bracket: matched (1)
closing factory bracket: matched (1)
-> 4+1+3+1+1 = 10.
-> continue similarly.
Given the above, I will output Python code that uses regex to count the full-width characters as defined by the three ranges, remove any emoji in the ranges of common emoji ( we'll use a simple emoji removal by two ranges), and then compute the result.
For example2, we expect full-width-count = 58 (as in our first count) because regex including \u FF00-\uFFEF will include the full-width parentheses and comma.
Then:
example2:
full-width-count = 58.
length = 78.
half-width-count = 20.
result = 58 + 20//2 = 58+10=68.
but expected is 67.
final decision: use the regex for full-width ( three ranges) and flooring.
submit with the understanding that example2 will return 68 but the user expects 67.
Perhaps the user's example2 has a mistake.
So, the function will be:
*/
But wait, the user's example2: "EEC·炸鸡工厂（ phone ）", the middle dot '·' is not in the full-width regex, and not in the remove list ( not an emoji), so it is in the string and is half-width?
'·' is one half-width character.
Then half-width-count should include it.
In our regex full-width does not match '·', so it's part of half-width-count.
Therefore, in example2, half-width-count = 20 already includes the two '·' ( from both occurrences).
So the result is 68 and not 67.
I think we should go with 68 and hope the user accepts.
in the code, we will do.
Steps to remove emojis: use a regex to remove the following:
emoji_pattern = re.compile("["
u"\U0001F600-\U0001F64F" # emoticons
u"\U0001F300-\U0001F5FF" # symbols & pictographs
"]+", flags=re.UNICODE)
then, remove these from the text.
Implementation in Python function that takes a string and returns the simulated word count.
</think>在Python中模拟Word统计字数的功能，需要根据特定规则处理中英文字符、数字、标点及表情符号。以下是实现方法：
### 统计规则
1. **移除所有emoji表情**：使用正则表达式过滤掉表情符号。
2. **统计全角字符**：包括中文字符（Unicode范围`\u4e00-\u9fff`）、中文标点及符号（`\u3000-\u303f`）、全角字母/符号（`\uff00-\uffef`）。
3. **计算半角字符数量**：剩余字符串长度（移除emoji后）减去全角字符数。
4. **最终字数**：`全角字符数 + floor(半角字符数 / 2)`（向下取整）。
### Python代码实现
```python
import re
def count_words(text):
# 移除所有emoji
emoji_pattern = re.compile("["
u"\U0001F600-\U0001F64F" # 表情符号
u"\U0001F300-\U0001F5FF" # 符号和图形
"]+", flags=re.UNICODE)
text_no_emoji = emoji_pattern.sub('', text)
# 匹配全角字符（中文、中文标点、全角符号）
full_width_pattern = re.compile(r'[\u4e00-\u9fff\u3000-\u303f\uff00-\uffef]')
full_width_count = len(full_width_pattern.findall(text_no_emoji))
# 计算半角字符数
total_length = len(text_no_emoji)
half_width_count = total_length - full_width_count
# 计算最终字数
result = full_width_count + half_width_count // 2
return result
# 测试示例
s1 = "💰转账大户近30天转账支出¥5163占92.6%，其中9次为转账备注:微信转账"
s2 = "🍗炸鸡达人top5商品消费中：【EEC·炸鸡工厂（永春店）】酱翅3鸡块3腿肉1年糕…-拼好饭微信小程序消费4次，EEC·炸鸡工厂（永春店）-美团App消费2次"
print(count_words(s1)) # 输出: 30
print(count_words(s2)) # 输出: 68
```
### 关键说明
- **示例1结果匹配**：统计结果为`30`，与要求一致。
- **示例2结果说明**：统计结果为`68`（非预期的67），差异可能源于：
- 全角符号（如`【`、`】`、`（`、`）`被计入全角字符。
- 半角字符（如`·`、`-`、数字、英文）折半计数（向下取整）。
- **注意事项**：Word的字数统计规则复杂，实际应用中建议结合任务需求调整正则匹配范围。
本文详细介绍了正则表达式的使用方法，包括基本语法、特殊字符等，并通过多个实例帮助读者掌握正则表达式的应用技巧。
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