非常实用的 Python 技巧

函数连续调用

def add(x):
    class AddNum(int):
        def __call__(self, x):
            return AddNum(self.numerator + x)
    return AddNum(x)

print add(2)(3)(5)
# 10
print add(2)(3)(4)(5)(6)(7)
# 27

# javascript 版
var add = function(x){
    var addNum = function(x){
        return add(addNum + x);
    };

    addNum.toString = function(){
        return x;
    }
    return addNum;
}

add(2)(3)(5)//10
add(2)(3)(4)(5)(6)(7)//27

默认值陷阱

>>> def evil(v=[]):
...     v.append(1)
...     print v
...
>>> evil()
[1]

>>> evil()
[1, 1]

读写csv文件

import csv

with open('data.csv', 'rb') as f:
    reader = csv.reader(f)
    for row in reader:
        print row

# 向csv文件写入
import csv

with open( 'data.csv', 'wb') as f:
    writer = csv.writer(f)
    writer.writerow(['name', 'address', 'age'])  # 单行写入
    data = [
            ( 'xiaoming ','china','10'),
            ( 'Lily', 'USA', '12')]

    writer.writerows(data)  # 多行写入

数制转换

>>> int('1000', 2)
8

>>> int('A', 16)
10

格式化 json

echo'{"k": "v"}' | python-m json.tool

list 扁平化

list_ = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
[k for i in list_ for k in i] #[1, 2, 3, 4, 5, 6, 7, 8, 9]
import numpy as np
print np.r_[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

import itertools
print list(itertools.chain(*[[1, 2, 3], [4, 5, 6], [7, 8, 9]]))
sum(list_, [])
flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
flatten(list_)

list 合并

>>> a = [1, 3, 5, 7, 9]
>>> b = [2, 3, 4, 5, 6]
>>> c = [5, 6, 7, 8, 9]
>>> list(set().union(a, b, c))
[1, 2, 3, 4, 5, 6, 7, 8, 9]

出现次数最多的 2 个字母

from collections import Counter
c = Counter('hello world')
print(c.most_common(2)) #[('l', 3), ('o', 2)]

谨慎使用

eval("__import__('os').system('rm -rf /')", {})

置换矩阵

matrix = [[1, 2, 3],[4, 5, 6]]
res = zip( *matrix )   # res = [(1, 4), (2, 5), (3, 6)]

列表推导

[item**2 for item in lst if item % 2]
map(lambda item: item ** 2, filter(lambda item: item % 2, lst))
>>> list(map(str, [1, 2, 3, 4, 5, 6, 7, 8, 9]))
['1', '2', '3', '4', '5', '6', '7', '8', '9']

排列组合

>>> for p in itertools.permutations([1, 2, 3, 4]):
...     print ''.join(str(x) for x in p)
...
1234
1243
1324
1342
1423
1432
2134
2143
2314
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4213
4231
4312
4321

>>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):
...     print ''.join(str(x) for x in c)
...
123
124
125
134
135
145
234
235
245
345
>>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):
...     print ''.join(str(x) for x in c)
...
11
12
13
22
23
33
>>> for p in itertools.product([1, 2, 3], [4, 5]):
(1, 4)
(1, 5)
(2, 4)
(2, 5)
(3, 4)
(3, 5)

默认字典

>>> m = dict()
>>> m['a']
Traceback (most recent call last):
  File "", line 1, in 
KeyError: 'a'
>>>
>>> m = collections.defaultdict(int)
>>> m['a']
0
>>> m['b']
0
>>> m = collections.defaultdict(str)
>>> m['a']
''
>>> m['b'] += 'a'
>>> m['b']
'a'
>>> m = collections.defaultdict(lambda: '[default value]')
>>> m['a']
'[default value]'
>>> m['b']
'[default value]'

反转字典

>>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
>>> m
{'d': 4, 'a': 1, 'b': 2, 'c': 3}
>>> {v: k for k, v in m.items()}
{1: 'a', 2: 'b', 3: 'c', 4: 'd'}

更多

https://www.zhihu.com/question/27376156
stackoverflow.com/questions/101268/hidden-features-of-python
https://zhuanlan.zhihu.com/p/22909144
https://www.zhihu.com/question/29995881