Question

我需要来自字符串

i = "1,'Test','items (one, two, etc.)',1,'long, list'"

提取下一个字符串的数组：

['1', "'Test'", "'items (one, two, etc.)'", '1', "'long, list'"]

借助regexpress

r=re.split(r',+(?=[^()]*(?:\(|$))', i)

我只收到下一个结果：

['1', "'Test'", "'items (one, two, etc.)'", '1', "'long", " list'"]

UPD1

应支持

NULL

i = "1,'Test',NULL,'items (one, two, etc.)',1,'long, list'"
['1', "'Test'", 'NULL', "'items (one, two, etc.)'", '1', "'long, list'"]

Answer 1

在这种情况下，您不需要re.split。您可以在列表理解中使用re.findall：

>>> [k for j in re.findall(r"(\d)|'([^']*)'",i) for k in j if k]
['1', 'Test', 'items (one, two, etc.)', '1', 'long, list']

前面的正则表达式会匹配一个引用'([^']*)'或任何数字（\d）之间的任何内容。

或者作为一种更有效的方式，您可以使用ast.literal_eval：

>>> from ast import literal_eval
>>> literal_eval(i)
(1, 'Test', 'items (one, two, etc.)', 1, 'long, list')

Answer 2

这是csv模块的任务：

import csv
from StringIO import StringIO
line = "1,'Test','items (one, two, etc.)',1,'long, list'"
reader = csv.reader(StringIO(line), quotechar="'")
row = next(reader)

# row == ['1', 'Test', 'items (one, two, etc.)', '1', 'long, list']

这里的关键是创建一个CSV阅读器，将单引号指定为引号字符。

Answer 3

您可以拆分单引号：

SELECT SUM(score) AS ttl_score, student_name 
FROM assessment 
WHERE class='$class_info' 
AND date>='$start' AND date<='$end' 
GROUP BY student_id ORDER BY ttl_score DESC;

或者使用地图：

i = "1,'Test','items (one, two, etc.)',1,'long, list'"



print([ele.strip(" ,") for ele in i.split("'") if ele.strip(",")])
['1', 'Test', 'items (one, two, etc.)', '1', 'long, list']

使用map with python 3非常有效：

print([ele for ele in map(lambda x:  x.strip(", "), i.split("'")) if ele])

更好地使用python2和In [7]: i = "1,'Test','items (one, two, etc.)',1,'long, list'" In [8]: timeit [ele for ele in map(lambda x: x.strip(", "), i.split("'")) if ele] 1000000 loops, best of 3: 1.5 µs per loop In [9]: r = re.compile(r"(\d)|'([^']*)'") In [10]: timeit [k for j in r.findall(i) for k in j if k] 100000 loops, best of 3: 3.92 µs per loop：

itertools.imap

所有这些都返回相同的输出栏literal_eval，因为它将数字计算为整数：

In [9]: from itertools  import imap   
In [10]: timeit [ele for ele in imap(lambda x:  x.strip(", "), i.split("'")) if ele]
1000000 loops, best of 3: 871 ns per loop  

In [11]: r = re.compile(r"(\d)|'([^']*)'")
In [12]: timeit [k for j in r.findall(i) for k in j if k]
100000 loops, best of 3: 4.27 µs per loop

In [17]: from ast import literal_eval
In [18]: timeit literal_eval(i)
100000 loops, best of 3: 16.2 µs per loop

NUll系列没有什么不同：

In [19]: literal_eval(i)
Out[19]: (1, 'Test', 'items (one, two, etc.)', 1, 'long, list')

In [20]: [k for j in r.findall(i) for k in j if k]
Out[20]: ['1', 'Test', 'items (one, two, etc.)', '1', 'long, list']

In [21]: [ele for ele in imap(lambda x:  x.strip(", "), i.split("'")) if ele]Out[21]: ['1', 'Test', 'items (one, two, etc.)', '1', 'long, list']

使用嵌套的分隔符号

3 个答案: