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'}

非常有用的 Python 技巧

10

27

javascript 版

向csv文件写入