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python基础入门实例操作之Python技巧

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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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