python惯用方法整理（Pythonic 3）

最新推荐文章于 2025-09-12 15:58:40 发布

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最新推荐文章于 2025-09-12 15:58:40 发布

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原文链接：https://my.oschina.net/Ashkandi/blog/760597

本文介绍了Python中list、tuple和dict的基本使用技巧，包括创建、判断空、元素计数及扩展方法等，并演示了如何利用namedtuple提高代码可读性，最后通过实例展示了如何进行排列组合。

2019独角兽企业重金招聘Python工程师标准>>>

list / tuple / dict（I）

注意区分list和tuple

简化创建list/tuple

# list or tuple
[ o.dosomething for o in objects if  True ]

或者

[ dosomething if x... else ... for x in Xs]

生成浮点数list

range只支持int，对于浮点数建议是用numpy的arange

>>> import numpy as np
>>> np.arange(0,1,0.1)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

特别注意：用range()方法计算得出的list会存在浮点计算的问题，例如：

>>>[x * 0.1 for x in range(0, 10)]
...[0.0, 0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6000000000000001, 0.7000000000000001, 0.8, 0.9]

named tuple

For example, it is common to represent a point, for example as a tuple (x, y). This leads to code like the following:

pt1 = (1.0, 5.0)
pt2 = (2.5, 1.5)

from math import sqrt
line_length = sqrt((pt1[0]-pt2[0])**2 + (pt1[1]-pt2[1])**2)

Using a named tuple it becomes more readable:

from collections import namedtuple
Point = namedtuple('Point', 'x y')
pt1 = Point(1.0, 5.0)
pt2 = Point(2.5, 1.5)

from math import sqrt
line_length = sqrt((pt1.x-pt2.x)**2 + (pt1.y-pt2.y)**2)

还可以用来“反射”

判断list/tuple是否为空

直接判断其是否为 True（或False即可）这种做法更“Pythonic”，对于用len(a)==0来判断，不是被推荐的。查看Python文档说明

a = []  # a = ()
if a:
    print("List is not empty.")
else:
    print("List is empty.")

>>> if not (None or False or 0 or 0.0 or 0j or '' or () or [] or {}):
...    print('False')
False

list() OR [] ?

原文
[] is faster than list()，同理()和tuple()、{}和dict()

>>> timeit.timeit('[]', number=10**7)
0.400970278520802
>>> timeit.timeit('list()', number=10 ** 7)
1.977615857740119
>>> timeit.timeit('()', number=10**7)
0.32171692848272926
>>> timeit.timeit('tuple()', number=10**7)
1.4221802492102285

list的append 和 extend方法的区别

Appends object at end.

x = [1, 2, 3]
x.append([4, 5])
print (x)  # gives you: [1, 2, 3, [4, 5]]

Extends list by appending elements from the iterable.

x = [1, 2, 3]
x.extend([4, 5])
print (x)  # gives you: [1, 2, 3, 4, 5]

获取list中某元素的数量

>>> [1, 2, 3, 4, 1, 4, 1].count(1)
3

If you are using Python 2.7 or 3 and you want number of occurrences for each element:

>>> from collections import Counter
>>> z = ['blue', 'red', 'blue', 'yellow', 'blue', 'red']
>>> Counter(z)
Counter({'blue': 3, 'red': 2, 'yellow': 1})

排列组合list中的元素

import itertools

Permutation (order matters):

>>> print (list(itertools.permutations([1,2,3,4], 2)))
[(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)]

Combination (order does NOT matter):

>>> print(list(itertools.combinations('123', 2)))
[('1', '2'), ('1', '3'), ('2', '3')]

Cartesian product (with several iterables):

>>> print(list(itertools.product([1,2,3], [4,5,6])))
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]

Cartesian product (with one iterable and itself):

>>> print(list(itertools.product([1,2], repeat=3)))
[(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)]

转载于:https://my.oschina.net/Ashkandi/blog/760597