python中的flatten用法_非常有用的 Python 技巧-优快云博客

来自公众号：苏生不惑

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)# 10print 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 csvwith open('data.csv', 'rb') as f: reader = csv.reader(f)for row in reader:print row# 向csv文件写入import csvwith 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 npprint np.r_[[1, 2, 3], [4, 5, 6], [7, 8, 9]]import itertoolsprint 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) ...123412431324134214231432213421432314234124132431312431423214324134123421412341324213423143124321>>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):... print ''.join(str(x) for x in c) ...123124125134135145234235245345>>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):... print ''.join(str(x) for x in c) ...111213222333>>> 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]'