python使用笔记

1 list加法

print([]+["55"])
# ['55']

2 dict用dict.get()方法添加元素以及添加list类型元素

dict.get()方法一般用来返回指定键的值，参考https://www.runoob.com/python/att-dictionary-get.html

# dict.get()方法 是用来返回指定键的值
# dict.get(key[, value]) 
# key -- 字典中要查找的键。
# value -- 可选，如果指定键的值不存在时，返回该默认值value。

tinydict = {'Name': 'Runoob', 'Age': 27}
print ("Age : %s" %  tinydict.get('Age'))
# Age : 27

# 没有设置 Sex，也没有设置默认的值，输出 None，反正不会报错
print ("Sex : %s" %  tinydict.get('Sex'))  
# Sex : None

# 没有设置 Salary，输出设置的值  0.0
print ('Salary: %s' % tinydict.get('Salary', 0.0))
# Salary: 0.0

# get() 方法 和  dict[key] 访问元素区别
# get(key) 方法在 key（键）不在字典中时，可以返回默认值 None 或者设置的默认值。
# dict[key] 在 key（键）不在字典中时，会触发 KeyError 异常。

dict.get()方法 也可以用来 给 dict 添加元素

history = {}
dic_1 = {"Name": "HaiCoder", "Age": 100, "Score": 99.5}

for name, metric in dic_1.items():
    history.get(name, [])
print(history)
print("---------------------")

for name, metric in dic_1.items():
    history[name] = history.get(name, [])
print(history)
print("---------------------")

for name, metric in dic_1.items():
    history[name] = history.get(name, [metric])
print(history)
print("---------------------")

for name, metric in dic_1.items():
    history[name] = history.get(name, []) + [metric]
print(history)
print("---------------------")

for name, metric in dic_1.items():
    history[name]  # 语法错误，不成立

# 返回如下
{}
---------------------
{'Name': [], 'Age': [], 'Score': []}
---------------------
{'Name': [], 'Age': [], 'Score': []}
---------------------
{'Name': ['HaiCoder'], 'Age': [100], 'Score': [99.5]}
---------------------

以下为进阶版

import pandas as pd

history = {}
dic_1 = {"Name": "HaiCoder", "Age": 100, "sex":"girl", "Score":99.5}
epochs = 10

for epoch in range(1, epochs + 1):
    dic_1["epoch"] = epoch
    for key, value in dic_1.items():
        history[key] = history.get(key, []) + [value]

print(history)
print("----------------------------------------")
print(pd.DataFrame(history))


# 结果为
{'Name': ['HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder'], 'Age': [100, 100, 100, 100, 100, 100, 100, 100, 100, 100], 'sex': ['girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl'], 'Score': [99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5], 'epoch': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]}
----------------------------------------
       Name  Age   sex  Score  epoch
0  HaiCoder  100  girl   99.5      1
1  HaiCoder  100  girl   99.5      2
2  HaiCoder  100  girl   99.5      3
3  HaiCoder  100  girl   99.5      4
4  HaiCoder  100  girl   99.5      5
5  HaiCoder  100  girl   99.5      6
6  HaiCoder  100  girl   99.5      7
7  HaiCoder  100  girl   99.5      8
8  HaiCoder  100  girl   99.5      9
9  HaiCoder  100  girl   99.5     10

这么做的原因：列表list中元素是顺序的，但是字典dict则是乱序的。当然也可以用collections.OrderedDict() 顺序字典。参考：https://blog.youkuaiyun.com/qq_42184799/article/details/86311804

3 OrderedDict有序字典使用及其和Dict的对比

注意：OrderedDict 和 Dict 的用法一致，甚至Dict也是有序的，只是为了保险，使用OrderedDict

import collections

dic_1 = {"Name": "HaiCoder", "Age": 100, "sex":"girl"}  # 
dic_2 = {"Score":99.5}
print(dic_1)
print("----------------------------------------")

print(dic_1.items())
print("----------------------------------------")

for k in dic_1.items():
    print(k)  # 对字典的遍历
    print(type(k))
print("----------------------------------------")

for k in dic_1:
    print(k)  # 对有序字典的每次遍历，结果都一样
    print(type(k))
print("----------------------------------------")

d3 = collections.OrderedDict(dic_2, **dic_1)
print(d3)
print("----------------------------------------")

print(d3.items())
print("----------------------------------------")

d4 = collections.OrderedDict(dic_1, **dic_2)
print(d4)
print("----------------------------------------")

d5 = collections.OrderedDict(dic_1)
print(d5)
print("----------------------------------------")

for k in d3.items():
    print(k)  # 对有序字典的每次遍历，结果都一样
    print(type(k))
print("----------------------------------------")

for k in d3:
    print(k)  # 对有序字典的每次遍历，结果都一样
    print(type(k))
print("----------------------------------------")

for k,v in d3.items():
    print (k,v)  # 对有序字典的每次遍历，结果都一样
print("----------------------------------------")

for k,v in d4.items():
    print (k,v)  # 对有序字典的每次遍历，结果都一样


# 返回
{'Name': 'HaiCoder', 'Age': 100, 'sex': 'girl'}
----------------------------------------
dict_items([('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl')])
----------------------------------------
('Name', 'HaiCoder')
<class 'tuple'>
('Age', 100)
<class 'tuple'>
('sex', 'girl')
<class 'tuple'>
----------------------------------------
Name
<class 'str'>
Age
<class 'str'>
sex
<class 'str'>
----------------------------------------
OrderedDict([('Score', 99.5), ('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl')])
----------------------------------------
odict_items([('Score', 99.5), ('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl')])
----------------------------------------
OrderedDict([('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl'), ('Score', 99.5)])
----------------------------------------
OrderedDict([('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl')])
----------------------------------------
('Score', 99.5)
<class 'tuple'>
('Name', 'HaiCoder')
<class 'tuple'>
('Age', 100)
<class 'tuple'>
('sex', 'girl')
<class 'tuple'>
----------------------------------------
Score
<class 'str'>
Name
<class 'str'>
Age
<class 'str'>
sex
<class 'str'>
----------------------------------------
Score 99.5
Name HaiCoder
Age 100
sex girl
----------------------------------------
Name HaiCoder
Age 100
sex girl
Score 99.5

进阶版，，，

import pandas as pd
import collections

history = collections.OrderedDict()
dic_1 = collections.OrderedDict({"Name": "HaiCoder", "Age": 100, "sex":"girl", "Score":99.5})

i = 0
dic_1["epoch"] = i

for key, value in dic_1.items():
    history[key] = history.get(key, []) + [value]
    i = i + 1
print(history)
print("----------------------------------------")

print(pd.DataFrame(history))
print("----------------------------------------")

epochs = 10

for epoch in range(1, epochs + 1):
    dic_1["epoch"] = epoch
    for key, value in dic_1.items():
        history[key] = history.get(key, []) + [value]

print(history)
print("----------------------------------------")

print(pd.DataFrame(history))


# 返回
OrderedDict([('Name', ['HaiCoder']), ('Age', [100]), ('sex', ['girl']), ('Score', [99.5]), ('epoch', [0])])
----------------------------------------
       Name  Age   sex  Score  epoch
0  HaiCoder  100  girl   99.5      0
----------------------------------------
OrderedDict([('Name', ['HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder', 'HaiCoder']), ('Age', [100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100]), ('sex', ['girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl', 'girl']), ('Score', [99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5, 99.5]), ('epoch', [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])])
----------------------------------------
        Name  Age   sex  Score  epoch
0   HaiCoder  100  girl   99.5      0
1   HaiCoder  100  girl   99.5      1
2   HaiCoder  100  girl   99.5      2
3   HaiCoder  100  girl   99.5      3
4   HaiCoder  100  girl   99.5      4
5   HaiCoder  100  girl   99.5      5
6   HaiCoder  100  girl   99.5      6
7   HaiCoder  100  girl   99.5      7
8   HaiCoder  100  girl   99.5      8
9   HaiCoder  100  girl   99.5      9
10  HaiCoder  100  girl   99.5     10

注意下面的报错

import pandas as pd
import collections

history = collections.OrderedDict()
dic_1 = collections.OrderedDict({"Name": "HaiCoder", "Age": 100, "sex":"girl", "Score":99.5})

i = 0

for key, value in dic_1.items():
    history["epoch"] = i  # 这个操作导致history字典中"epoch"键对应的值是数字类型int
    history[key] = history.get(key, []) + [value]  
    
print(history)
print("----------------------------------------")

print(pd.DataFrame(history))
print("----------------------------------------")

epochs = 10

for epoch in range(1, epochs + 1):
    dic_1["epoch"] = epoch  # 这个操作使得dic_1字典中"epoch"键对应的值是数字类型int
    for key, value in dic_1.items():
        history[key] = history.get(key, []) + [value]  # 但是这个操作history字典中"epoch"键对应的值是列表list类型，会与前面已经定义好的history字典中已经存在的epoch"键对应的int类型值起冲突

print(history)
print("----------------------------------------")

print(pd.DataFrame(history))


# 返回
OrderedDict([('epoch', 0), ('Name', ['HaiCoder']), ('Age', [100]), ('sex', ['girl']), ('Score', [99.5])])
----------------------------------------
   epoch      Name  Age   sex  Score
0      0  HaiCoder  100  girl   99.5
----------------------------------------
Traceback (most recent call last):
  File "D:/Resource_Learning/LIWEI/Example.py", line 183, in <module>
    history[key] = history.get(key, []) + [value]
TypeError: unsupported operand type(s) for +: 'int' and 'list'


# 报错原因
因为前面  history["epoch"] = i  # 这个操作导致history字典中"epoch"键对应的值是数字类型int
而后面  history[key] = history.get(key, []) + [value]  # 但是这个操作history字典中"epoch"键对应的值是列表list类型，会与前面已经定义好的history字典中已经存在的epoch"键对应的int类型值起冲突

这个报错在dict中一样会出现

OrderedDict和Dict也可以混用，所以对排序精度要求没那么严格，就直接用dict()吧

import collections

history = collections.OrderedDict()
dic_1 = collections.OrderedDict({"Name": "HaiCoder", "Age": 100, "sex":"girl"})
dic_2 = {"Score":99.5}
dic_3 = collections.OrderedDict({"Score":99.5})

print(collections.OrderedDict(dic_1,**dic_2))
print("----------------------------------------")

print(dict(dic_1,**dic_2))
print("----------------------------------------")

print(dict(dic_1,**dic_3))
print("----------------------------------------")

print(collections.OrderedDict(dic_1,**dic_3))
print("----------------------------------------")


# 返回
OrderedDict([('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl'), ('Score', 99.5)])
----------------------------------------
{'Name': 'HaiCoder', 'Age': 100, 'sex': 'girl', 'Score': 99.5}
----------------------------------------
{'Name': 'HaiCoder', 'Age': 100, 'sex': 'girl', 'Score': 99.5}
----------------------------------------
OrderedDict([('Name', 'HaiCoder'), ('Age', 100), ('sex', 'girl'), ('Score', 99.5)])
----------------------------------------