流畅的python_第二章

最新推荐文章于 2023-11-01 21:43:43 发布

翻译最新推荐文章于 2023-11-01 21:43:43 发布 · 509 阅读

顺序我也忘记了，就随便弄了

切片：

# 普通的切片
>>> l = [10, 20, 30, 40, 50, 60]
>>> l[:2] # 在下标2的地方分割
[10, 20]
>>> l[2:]
[30, 40, 50, 60]
>>> l[:3] # 在下标3的地方分割
[10, 20, 30]
>>> l[3:]
[40, 50, 60]

# 对对象进行切片
>>> s = 'bicycle'
>>> s[::3]
'bye'
>>> s[::-1]
'elcycib'
>>> s[::-2]
'eccb'

# slice(a, b, c)
>>> invoice = """
... 0.....6................................40........52...55........
... 1909 Pimoroni PiBrella                 $17.50    3    $52.50
... 1489 6mm Tactile Switch x20            $4.95     2    $9.90
... 1510 Panavise Jr. - PV-201             $28.00    1    $28.00
... 1601 PiTFT Mini Kit 320x240 $34.95 1   $34.95
... """
>>> SKU = slice(0, 6)
>>> DESCRIPTION = slice(6, 40)
>>> UNIT_PRICE = slice(40, 52)
>>> QUANTITY = slice(52, 55)
>>> ITEM_TOTAL = slice(55, None)
>>> line_items = invoice.split('\n')[2:]
>>> for item in line_items:
...     print(item[UNIT_PRICE], item[DESCRIPTION])
...
    $17.50  Pimoroni PiBrella
    $4.95   6mm Tactile Switch x20
    $28.00  Panavise Jr. - PV-201
    $34.95  PiTFT Mini Kit 320x240

# 给切片赋值
>>> l = list(range(10))
>>> l
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> l[2:5] = [20, 30]
>>> l
[0, 1, 20, 30, 5, 6, 7, 8, 9]
>>> del l[5:7]
>>> l
[0, 1, 20, 30, 5, 8, 9]
>>> l[3::2] = [11, 22]
>>> l
[0, 1, 20, 11, 5, 22, 9]
>>> l[2:5] = 100 
Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
TypeError: can only assign an iterable
>>> l[2:5] = [100]
>>> l
[0, 1, 100, 22, 9]

素组：

# 一个浮点型数组的创建、存入文件和从文件读取的过程
>>> from array import array
>>> from random import random
>>> floats = array('d', (random() for i in range(10**7)))
>>> floats[-1]
0.07802343889111107
>>> fp = open('floats.bin', 'wb')
>>> floats.tofile(fp)
>>> fp.close()
>>> floats2 = array('d')
>>> fp = open('floats.bin', 'rb')
>>> floats2.fromfile(fp, 10**7)
>>> fp.close()
>>> floats2[-1]
0.07802343889111107
>>> floats2 == floats
True

# 对numpy.ndarray 的行和列进行基本操作
>>> import numpy
>>> a = numpy.arange(12)
>>> a
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
>>> type(a)
<class 'numpy.ndarray'>
>>> a.shape
(12,)
>>> a.shape = 3, 4
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> a[2]
array([ 8, 9, 10, 11])
>>> a[2, 1]
9
>>> a[:, 1]
array([1, 5, 9])
>>> a.transpose()
array([[ 0, 4, 8],
[ 1, 5, 9],
[ 2, 6, 10],
[ 3, 7, 11]])

# 可以对numpy.ndarray 中的元素进行抽象的读取、保存和其他操作：
>>> import numpy
>>> floats = numpy.loadtxt('floats-10M-lines.txt')
>>> floats[-3:]
array([ 3016362.69195522, 535281.10514262, 4566560.44373946])
>>> floats *= .5
>>> floats[-3:]
array([ 1508181.34597761, 267640.55257131, 2283280.22186973])
>>> from time import perf_counter as pc
>>> t0 = pc(); floats /= 3; pc() - t0
0.03690556302899495
>>> numpy.save('floats-10M', floats)
>>> floats2 = numpy.load('floats-10M.npy', 'r+')
>>> floats2 *= 6
>>> floats2[-3:]
memmap([3016362.69195522, 535281.10514262, 4566560.44373946])

# 双向队列
>>> from collections import deque
>>> dq = deque(range(10), maxlen=10)
>>> dq
deque([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], maxlen=10)
>>> dq.rotate(3)
>>> dq
deque([7, 8, 9, 0, 1, 2, 3, 4, 5, 6], maxlen=10)
>>> dq.rotate(-4)
>>> dq
deque([1, 2, 3, 4, 5, 6, 7, 8, 9, 0], maxlen=10)
>>> dq.appendleft(-1)
>>> dq
deque([-1, 1, 2, 3, 4, 5, 6, 7, 8, 9], maxlen=10)
>>> dq.extend([11, 22, 33])
>>> dq
deque([3, 4, 5, 6, 7, 8, 9, 11, 22, 33], maxlen=10)
>>> dq.extendleft([10, 20, 30, 40])
>>> dq
deque([40, 30, 20, 10, 3, 4, 5, 6, 7, 8], maxlen=10)

序列：

# 对序列使用+和*
>>> l = [1, 2, 3]
>>> l * 5
[1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3]
>>> 5 * 'abcd'
'abcdabcdabcdabcdabcd'

>>> board = [['_'] * 3 for i in range(3)] 
>>> board
[['_', '_', '_'], ['_', '_', '_'], ['_', '_', '_']]
>>> board[1][2] = 'X' 
>>> board
[['_', '_', '_'], ['_', '_', 'X'], ['_', '_', '_']]

# 含有3 个指向同一对象的引用的列表是毫无用处的
>>> weird_board = [['_'] * 3] * 3
>>> weird_board
[['_', '_', '_'], ['_', '_', '_'], ['_', '_', '_']]
>>> weird_board[1][2] = 'O'
>>> weird_board
[['_', '_', 'O'], ['_', '_', 'O'], ['_', '_', 'O']]

# 序列的增量赋值
>>> l = [1, 2, 3]
>>> id(l)
4311953800
>>> l *= 2
>>> l
[1, 2, 3, 1, 2, 3]
>>> id(l)
4311953800
>>> t = (1, 2, 3)
>>> id(t)
4312681568
>>> t *= 2
>>> id(t)
4301348296

# 一个关于+=的谜题
>>> t = (1, 2, [30, 40])
>>> t[2] += [50, 60]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment
>>> t
(1, 2, [30, 40, 50, 60])

列表生产式与生成器：

symbols = '$¢£¥€¤'
codes = [ord(symbol) for symbol in symbols]
print(codes)


# 生成器表达式的语法跟列表推导差不多，只不过把方括号换成圆括号而已。

# 普通模式
>>> symbols = '$¢£¥€¤'
>>> codes = []
>>> for symbol in symbols:
... codes.append(ord(symbol))
...
>>> codes
[36, 162, 163, 165, 8364, 164]

# 推导式
>>> symbols = '$¢£¥€¤'
>>> codes = [ord(symbol) for symbol in symbols]
>>> codes
[36, 162, 163, 165, 8364, 164]

# 列表推导和map/filter 组合来创建同样的表单
>>> symbols = '$¢£¥€¤'
>>> beyond_ascii = [ord(s) for s in symbols if ord(s) > 127]
>>> beyond_ascii
[162, 163, 165, 8364, 164]
>>> beyond_ascii = list(filter(lambda c: c > 127, map(ord, symbols)))
>>> beyond_ascii
[162, 163, 165, 8364, 164]

# 笛卡儿积
>>> colors = ['black', 'white']
>>> sizes = ['S', 'M', 'L']
>>> tshirts = [(color, size) for color in colors for size in sizes]
>>> tshirts
[('black', 'S'), ('black', 'M'), ('black', 'L'), ('white', 'S'),
('white', 'M'), ('white', 'L')]
>>> for color in colors:
...     for size in sizes:
...         print((color, size))
...
('black', 'S')
('black', 'M')
('black', 'L')
('white', 'S')
('white', 'M')
('white', 'L')
>>> tshirts = [(color, size) for size in sizes
...                          for color in colors]
>>> tshirts
[('black', 'S'), ('white', 'S'), ('black', 'M'), ('white', 'M'),
('black', 'L'), ('white', 'L')]

# 生成器表达式
# 用生成器表达式初始化元组和数组
>>> symbols = '$¢£¥€¤'
>>> tuple(ord(symbol) for symbol in symbols)
(36, 162, 163, 165, 8364, 164)
>>> import array
>>> array.array('I', (ord(symbol) for symbol in symbols))
array('I', [36, 162, 163, 165, 8364, 164])

使用生成器表达式计算笛卡儿积
>>> colors = ['black', 'white']
>>> sizes = ['S', 'M', 'L']
>>> for tshirt in ('%s %s' % (c, s) for c in colors for s in sizes):
... print(tshirt)
...
black S
black M
black L
white S
white M
white L

元组：

# 把元组用作记录
>>> lax_coordinates = (33.9425, -118.408056)
>>> city, year, pop, chg, area = ('Tokyo', 2003, 32450, 0.66, 8014)
>>> traveler_ids = [('USA', '31195855'), ('BRA', 'CE342567'),
... ('ESP', 'XDA205856')]
>>> for passport in sorted(traveler_ids):
...     print('%s/%s' % passport)
...
BRA/CE342567
ESP/XDA205856
USA/31195855
>>> for country, _ in traveler_ids:
...     print(country)
...
USA
BRA
ESP

# 元组拆包
>>> lax_coordinates = (33.9425, -118.408056)
>>> latitude, longitude = lax_coordinates # 元组拆包
>>> latitude
33.9425
>>> longitude
-118.408056

# * 运算符把一个可迭代对象拆开作为函数的参数：
>>> divmod(20, 8)
(2, 4)
>>> t = (20, 8)
>>> divmod(*t)
(2, 4)
>>> quotient, remainder = divmod(*t)
>>> quotient, remainder
(2, 4)

>>> import os
>>> _, filename = os.path.split('/home/luciano/.ssh/idrsa.pub')
>>> filename
'idrsa.pub'
>>> a, b, *rest = range(5)
>>> a, b, rest
(0, 1, [2, 3, 4])
>>> a, b, *rest = range(3)
>>> a, b, rest
(0, 1, [2])
>>> a, b, *rest = range(2)
>>> a, b, rest
(0, 1, [])

>>> a, *body, c, d = range(5)
>>> a, body, c, d
(0, [1, 2], 3, 4)
>>> *head, b, c, d = range(5)
>>> head, b, c, d
([0, 1], 2, 3, 4)

# 用嵌套元组来获取经度
metro_areas = [
    ('Tokyo','JP',36.933,(35.689722,139.691667)), 
    ('Delhi NCR', 'IN', 21.935, (28.613889, 77.208889)),
    ('Mexico City', 'MX', 20.142, (19.433333, -99.133333)),
    ('New York-Newark', 'US', 20.104, (40.808611, -74.020386)),
    ('Sao Paulo', 'BR', 19.649, (-23.547778, -46.635833)),
]
print('{:15} | {:^9} | {:^9}'.format('', 'lat.', 'long.'))
fmt = '{:15} | {:9.4f} | {:9.4f}'
for name, cc, pop, (latitude, longitude) in metro_areas: 
    if longitude <= 0: 
        print(fmt.format(name, latitude, longitude))

                | lat.      | long.
Mexico City     | 19.4333   | -99.1333
New York-Newark | 40.8086   | -74.0204
Sao Paul        | -23.5478  | -46.6358

# 定义和使用具名元组
# 一个具名元组需要两个参数，一个是类名，另一个是类的各个字段的名字
>>> from collections import namedtuple
>>> City = namedtuple('City', 'name country population coordinates')
>>> tokyo = City('Tokyo', 'JP', 36.933, (35.689722, 139.691667))
>>> tokyo
City(name='Tokyo', country='JP', population=36.933, coordinates=(35.689722,
139.691667))
>>> tokyo.population
36.933
>>> tokyo.coordinates
(35.689722, 139.691667)
>>> tokyo[1]
'JP'

# 具名元组的属性和方法（接续前一个示例）
>>> City._fields 
('name', 'country', 'population', 'coordinates')
>>> LatLong = namedtuple('LatLong', 'lat long')
>>> delhi_data = ('Delhi NCR', 'IN', 21.935, LatLong(28.613889, 77.208889))
>>> delhi = City._make(delhi_data) 
>>> delhi._asdict() 
OrderedDict([('name', 'Delhi NCR'), ('country', 'IN'), ('population',
21.935), ('coordinates', LatLong(lat=28.613889, long=77.208889))])
>>> for key, value in delhi._asdict().items():
print(key + ':', value)
name: Delhi NCR
country: IN
population: 21.935
coordinates: LatLong(lat=28.613889, long=77.208889)

#元组是一种很强大的可以当作记录来用的数据类型。它的第二个角色则是充当一个不可变的列表

排序：

# sorted排序
>>> fruits = ['grape', 'raspberry', 'apple', 'banana']
>>> sorted(fruits)
['apple', 'banana', 'grape', 'raspberry']
>>> fruits
['grape', 'raspberry', 'apple', 'banana']
>>> sorted(fruits, reverse=True)
['raspberry', 'grape', 'banana', 'apple']
>>> sorted(fruits, key=len)
['grape', 'apple', 'banana', 'raspberry']
>>> sorted(fruits, key=len, reverse=True)
['raspberry', 'banana', 'grape', 'apple']
>>> fruits
['grape', 'raspberry', 'apple', 'banana']
>>> fruits.sort()
>>> fruits
['apple', 'banana', 'grape', 'raspberry']

# 在有序序列中用bisect 查找某个元素的插入位置
import bisect
import sys
HAYSTACK = [1, 4, 5, 6, 8, 12, 15, 20, 21, 23, 23, 26, 29, 30]
NEEDLES = [0, 1, 2, 5, 8, 10, 22, 23, 29, 30, 31]
ROW_FMT = '{0:2d} @ {1:2d} {2}{0:<2d}'
def demo(bisect_fn):
    for needle in reversed(NEEDLES):
        position = bisect_fn(HAYSTACK, needle)
        offset = position * ' |'
        print(ROW_FMT.format(needle, position, offset))
if __name__ == '__main__':
    if sys.argv[-1] == 'left':
        bisect_fn = bisect.bisect_left
    else:
        bisect_fn = bisect.bisect
    print('DEMO:', bisect_fn.__name__)
    print('haystack ->', ' '.join('%2d' % n for n in HAYSTACK))
    demo(bisect_fn)
    
    
# 根据一个分数，找到它所对应的成绩
>>> def grade(score, breakpoints=[60, 70, 80, 90], grades='FDCBA'):
... i = bisect.bisect(breakpoints, score)
... return grades[i]
...
>>> [grade(score) for score in [33, 99, 77, 70, 89, 90, 100]]
['F', 'A', 'C', 'C', 'B', 'A', 'A']

# insort 可以保持有序序列的顺序
import bisect
import random
SIZE=7
random.seed(1729)
my_list = []
for i in range(SIZE):
    new_item = random.randrange(SIZE*2)
    bisect.insort(my_list, new_item)
    print('%2d ->' % new_item, my_list)