[英]replacing part of a string with value from another column
A pandas DataFrame contains a column with descriptions and placeholders in curly braces: pandas DataFrame包含一个列,其中包含花括号中的描述和占位符:
descr replacement
This: {should be replaced} with this
The task is to replace the text in the curly braces with text from another column in the same row. 任务是用同一行中另一列的文本替换花括号中的文本。 It's unfortunately not as easy as: 遗憾的是,它不像以下那么容易:
df["descr"] = df["descr"].str.replace(r"{*?}", df["replacement"])
~/anaconda3/lib/python3.6/site-packages/pandas/core/strings.py in replace(self, pat, repl, n, case, flags, regex)
2532 def replace(self, pat, repl, n=-1, case=None, flags=0, regex=True):
2533 result = str_replace(self._parent, pat, repl, n=n, case=case,
-> 2534 flags=flags, regex=regex)
2535 return self._wrap_result(result)
2536
~/anaconda3/lib/python3.6/site-packages/pandas/core/strings.py in str_replace(arr, pat, repl, n, case, flags, regex)
548 # Check whether repl is valid (GH 13438, GH 15055)
549 if not (is_string_like(repl) or callable(repl)):
--> 550 raise TypeError("repl must be a string or callable")
551
552 is_compiled_re = is_re(pat)
TypeError: repl must be a string or callable
Use list comprehension with re.sub
, especially if performance is important: 将list comprehension与re.sub
,特别是如果性能很重要:
import re
df['new'] = [re.sub(r"{.*?}", b, a) for a, b in zip(df['descr'], df['replacement'])]
print (df)
descr replacement new
0 This: {should be replaced} with this This: with this
1 This: {data} aaa This: aaa
Your code is using the Pandas.Series.str.replace() and it expects two strings to perform the replacement operation, but the second parameter is a Series. 您的代码使用的是Pandas.Series.str.replace() ,它需要两个字符串来执行替换操作,但第二个参数是Series。
Series.str.replace(pat, repl, n=-1, case=None, flags=0, regex=True)[source] Series.str.replace(pat,repl,n = -1,case = None,flags = 0,regex = True)[来源]
Replace occurrences of pattern/regex in the Series/Index with some other string. 用一些其他字符串替换Series / Index中出现的pattern / regex。 Equivalent to str.replace() or re.sub(). 相当于str.replace()或re.sub()。 Parameters: 参数:
pat : string or compiled regex pat:字符串或编译的正则表达式
repl : string or callable ... repl:string或callable ...
You can correct it using directly the Pandas.Series.replace() method: 您可以直接使用Pandas.Series.replace()方法更正它:
df = pd.DataFrame({'descr': ['This: {should be replaced}'],
'replacement': 'with this'
})
>> df["descr"].replace(r"{.+?}", df["replacement"], regex = True)
0 This: with this
Observation: 观察:
I changed a bit of your regexp. 我改变了你的正则表达式。
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