非常有用的 Python 技巧

gclome

非常有用的 Python 技巧
函数连续调用

1. def add(x):
   
2.     class AddNum(int):
   
3.         def __call__(self, x):
   
4.             return AddNum(self.numerator + x)
   
5.     return AddNum(x)
   
6. 
   
7. print add(2)(3)(5)
   
8. # 10
   
9. print add(2)(3)(4)(5)(6)(7)
   
10. # 27
    
11. 
    
12. # javascript 版
    
13. var add = function(x){
    
14.     var addNum = function(x){
    
15.         return add(addNum + x);
    
16.     };
    
17. 
    
18.     addNum.toString = function(){
    
19.         return x;
    
20.     }
    
21.     return addNum;
    
22. }
    
23. 
    
24. add(2)(3)(5)//10
    
25. add(2)(3)(4)(5)(6)(7)//27

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默认值陷阱

1. >>> def evil(v=[]):
   
2. ...     v.append(1)
   
3. ...     print v
   
4. ...
   
5. >>> evil()
   
6. [1]
   
7. 
   
8. >>> evil()
   
9. [1, 1]

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读写csv文件

1. import csv
   
2. 
   
3. with open('data.csv', 'rb') as f:
   
4.     reader = csv.reader(f)
   
5.     for row in reader:
   
6.         print row
   
7. 
   
8. # 向csv文件写入
   
9. import csv
   
10. 
    
11. with open( 'data.csv', 'wb') as f:
    
12.     writer = csv.writer(f)
    
13.     writer.writerow(['name', 'address', 'age'])  # 单行写入
    
14.     data = [
    
15.             ( 'xiaoming ','china','10'),
    
16.             ( 'Lily', 'USA', '12')]
    
17. 
    
18.     writer.writerows(data)  # 多行写入

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数制转换

1. >>> int('1000', 2)
   
2. 8
   
3. 
   
4. >>> int('A', 16)
   
5. 10

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格式化 json

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

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list 扁平化

1. list_ = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
   
2. [k for i in list_ for k in i] #[1, 2, 3, 4, 5, 6, 7, 8, 9]
   
3. import numpy as np
   
4. print np.r_[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
   
5. 
   
6. import itertools
   
7. print list(itertools.chain(*[[1, 2, 3], [4, 5, 6], [7, 8, 9]]))
   
8. sum(list_, [])
   
9. flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
   
10. flatten(list_)

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list 合并

1. >>> a = [1, 3, 5, 7, 9]
   
2. >>> b = [2, 3, 4, 5, 6]
   
3. >>> c = [5, 6, 7, 8, 9]
   
4. >>> list(set().union(a, b, c))
   
5. [1, 2, 3, 4, 5, 6, 7, 8, 9]

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出现次数最多的 2 个字母

1. from collections import Counter
   
2. c = Counter('hello world')
   
3. print(c.most_common(2)) #[('l', 3), ('o', 2)]

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谨慎使用

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

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置换矩阵

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

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列表推导

1. [item**2 for item in lst if item % 2]
   
2. map(lambda item: item ** 2, filter(lambda item: item % 2, lst))
   
3. >>> list(map(str, [1, 2, 3, 4, 5, 6, 7, 8, 9]))
   
4. ['1', '2', '3', '4', '5', '6', '7', '8', '9']
   
5. 
   

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排列组合

1. >>> for p in itertools.permutations([1, 2, 3, 4]):
   
2. ...     print ''.join(str(x) for x in p)
   
3. ...
   
4. 1234
   
5. 1243
   
6. 1324
   
7. 1342
   
8. 1423
   
9. 1432
   
10. 2134
    
11. 2143
    
12. 2314
    
13. 2341
    
14. 2413
    
15. 2431
    
16. 3124
    
17. 3142
    
18. 3214
    
19. 3241
    
20. 3412
    
21. 3421
    
22. 4123
    
23. 4132
    
24. 4213
    
25. 4231
    
26. 4312
    
27. 4321
    
28. 
    
29. >>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):
    
30. ...     print ''.join(str(x) for x in c)
    
31. ...
    
32. 123
    
33. 124
    
34. 125
    
35. 134
    
36. 135
    
37. 145
    
38. 234
    
39. 235
    
40. 245
    
41. 345
    
42. >>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):
    
43. ...     print ''.join(str(x) for x in c)
    
44. ...
    
45. 11
    
46. 12
    
47. 13
    
48. 22
    
49. 23
    
50. 33
    
51. >>> for p in itertools.product([1, 2, 3], [4, 5]):
    
52. (1, 4)
    
53. (1, 5)
    
54. (2, 4)
    
55. (2, 5)
    
56. (3, 4)
    
57. (3, 5)

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默认字典

1. >>> m = dict()
   
2. >>> m['a']
   
3. Traceback (most recent call last):
   
4.   File "<stdin>", line 1, in <module>
   
5. KeyError: 'a'
   
6. >>>
   
7. >>> m = collections.defaultdict(int)
   
8. >>> m['a']
   
9. 0
   
10. >>> m['b']
    
11. 0
    
12. >>> m = collections.defaultdict(str)
    
13. >>> m['a']
    
14. ''
    
15. >>> m['b'] += 'a'
    
16. >>> m['b']
    
17. 'a'
    
18. >>> m = collections.defaultdict(lambda: '[default value]')
    
19. >>> m['a']
    
20. '[default value]'
    
21. >>> m['b']
    
22. '[default value]'

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反转字典

1. >>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
   
2. >>> m
   
3. {'d': 4, 'a': 1, 'b': 2, 'c': 3}
   
4. >>> {v: k for k, v in m.items()}
   
5. {1: 'a', 2: 'b', 3: 'c', 4: 'd'}

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