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Question

如果出现字符“ /devices/ered34wa/logs”，我如何最有效地切出单词的一部分，如果出现字符“ =#=”，则完成切单词的效果如何？例如：

通过大字符串

=#=

python代码返回：

'321@5=85@45@41=#=I-LOVE-STACK-OVER-FLOW=#=3234@41@=q#$^1=@=xx$q=@=xpa$=4319'

任何帮助将不胜感激。

Answer 1

使用split()：

s = '321@5=85@45@41=#=I-LOVE-STACK-OVER-FLOW=#=3234@41@=q#$^1=@=xx$q=@=xpa$=4319'

st = '=#='
ed = '=#='
print((s.split(st))[1].split(ed)[0])

使用regex：

import re
s = '321@5=85@45@41=#=I-LOVE-STACK-OVER-FLOW=#=3234@41@=q#$^1=@=xx$q=@=xpa$=4319'

print(re.search('%s(.*)%s' % (st, ed), s).group(1))

输出：

I-LOVE-STACK-OVER-FLOW

Answer 2

除了@DirtyBit的答案外，如果您还想处理2个以上'=＃='的情况，则可以拆分字符串，然后添加其他所有元素：

import pandas as pd

df = pd.read_excel('bb.xlsx')

jobs = set(df['Job'])      #remove duplicates
result = [[[ (df['M1'][i],0), (df['M2'][i],1), (df['M3'][i],2) ] for i in df.index if df['Job'][i] == job] for job in jobs]
print(result)

输出

s = '321@5=85@45@41=#=I-LOVE-STACK-OVER-FLOW=#=3234@41@=q#$^1=@=xx$q=@=xpa$=4319=#=|I-ALSO-LOVE-SO=#=3123123'
parts = s.split('=#=')
print(''.join([parts[i] for i in range(1,len(parts),2)]))

Answer 3

说明在代码中。

import re


ori_list = re.split("=#=",ori_str)
    # you can imagine your goal is to find the string wrapped between signs of "=#=" 
    # so after the split, the even number position must be the parts outsides of "=#=" 
    # and the odd number position is what you want
for i in range(len(ori_list)):
    if i%2 == 1:#odd position
       print(ori_list[i])

如果出现符号，请将单词（字符串）的一部分更改为其他字符串。蟒蛇

3 个答案:

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