[英]RegEx: match line breaks or spaces between square brackets
如何找到方括號內的所有換行符或空格?
我有以下字符串:
{
"decay_id": {
"int_only": true,
"feature_type": "categorical",
"category_values": [
0,
1
],
"category_names": [
"d1",
"d2"
]
}
}
我想刪除換行符和空格,以便我得到:
{
"decay_id": {
"int_only": true,
"feature_type": "categorical",
"category_values": [0,1],
"category_names": ["d1","d2"]
}
}
我如何使用 RegEx(和 Python)來做到這一點?
就像是:
import re
output = re.sub("some RegEx", "", input)
利用
import re
output = re.sub(r"\[[^][]*]", lambda z: re.sub(r'\s+', '', z.group()), input)
結果:
{
"decay_id": {
"int_only": true,
"feature_type": "categorical",
"category_values": [0,1],
"category_names": ["d1","d2"]
}
}
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