[英]How to replace all square brackets and their values with empty strings?
如果您不知道有多少個方括號,如何用空字符串替換所有方括號及其值?
data = "[a] 1 [b] test [c] other characters"
我希望數據是" 1 test other characters"
。
您可以使用正則表達式在括號部分拆分字符串:
import re
data = "[a] 1 [b] 2 [c]"
parts = re.split(r'\[.*?\]', data)
# ['', ' 1 ', ' 2 ', '']
out = '[' + ' '.join(parts) + ']'
print(out)
# [ 1 2 ]
另一個冗長的答案是:
import re
data = "[a] 1 [b] test [c] other characters"
pattern = '\[.*?\]' #any character(s) with [ ]
unwanted = set(re.findall(pattern,data))
results = ' '.join(i for i in data.split()
if i not in unwanted)
print(results)
# 1 test other characters
更新
加法條件
問題 > “[a] 1 [b] test [1] other [c]haracters” 由於.split()
這不會被替換
import re
data = "[a] 1 [b] test [1] other [c]haracters"
pattern = '\[.*?\]' #any character(s) with [ ]
print(re.sub(pattern,'',data))
# " 1 test other haracters"
刪除多余的空格
_ = re.sub(pattern,'',data).split()
print(' '.join(i for i in _ if i))
# "1 test other haracters"
嘗試正則表達式替代。
import re
data = "[a] 1 [b] 2 [c]"
new = re.sub("([[a-z]*])", "", data)
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