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)//10add(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 Counterc = 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 "<stdin>", line 1, in <module>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'}
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