函数连续调用

  1.  
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

  1.  
>>> def evil(v=[]):

...     v.append(1)

...     print v

...

>>> evil()

[1]

>>> evil()

[1, 1]


读写csv文件

  1.  
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

  1.  
  2. >>> int('1000', 2)

8

>>> int('A', 16)

10

格式化 json

echo'{"k": "v"}'|python-m json.tool



list 扁平化

  1.  
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

  1.  
>>> 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

  1.  
from collections import Counter

c = Counter('hello world')

print(c.most_common(2)) #[('l', 3), ('o', 2)]


谨慎使用

  1.  
eval（"__import__('os').system('rm -rf /')", {})


置换矩阵

  1.  
 matrix = [[1, 2, 3],[4, 5, 6]]

res = zip( *matrix )   # res = [(1, 4), (2, 5), (3, 6)]


列表推导

  1.  
[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

  1.  
>>> 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

  1.  
>>> 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

  1.  
>>> 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'}