Python新手要掌握的实用技巧
一、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/
二、默认值陷阱
>>> def evil(v=[]):
... v.append(1)
... print v
...
>>> evil()
[1]
>>> evil()
[1,
1]
三、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
四、数制转换
>>> int('1000',
2)
8
>>> int('A',
16)
10
五、格式化 json
echo'{"k": "v"}' | python-m json.tool
六、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_)
七、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]
八、出现次数较多的 2 个字母
from collections
import Counter
c = Counter('hello world')
print(c.most_common(2))
#[('l', 3), ('o', 2)]
九、谨慎使用
eval("__import__('os').system('rm -rf /')", {})
十、置换矩阵
matrix = [[1,
2,
3],[4,
5,
6]]
res = zip( *matrix )
# res = [(1, 4), (2, 5), (3, 6)]
十一、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']
十二、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)
十三、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]'
十四、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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