本篇内容介绍了“有哪些有用的Python技巧”的有关知识,在实际案例的操作过程中,不少人都会遇到这样的困境,接下来就让小编带领大家学习一下如何处理这些情况吧!希望大家仔细阅读,能够学有所成!
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 "<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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