pandas 通过pandas数据框用空格替换str列的换行符
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Replacing newlines with spaces for str columns through pandas dataframe
提问by alvas
Given an example dataframe with the 2nd and 3rd columns of free text, e.g.
给定一个带有自由文本第 2 和第 3 列的示例数据框，例如
>>> import pandas as pd
>>> lol = [[1,2,'abc','foo\nbar'], [3,1, 'def\nhaha', 'love it\n']]
>>> pd.DataFrame(lol)
0 1 2 3
0 1 2 abc foo\nbar
1 3 1 def\nhaha love it\n
The goal is to replace the \nto (whitespace) and strip the string in column 2 and 3 to achieve:
目标是替换\nto （空格）并去除第 2 列和第 3 列中的字符串以实现：
>>> pd.DataFrame(lol)
0 1 2 3
0 1 2 abc foo bar
1 3 1 def haha love it
How to replace newlines with spaces for specific columns through pandas dataframe?
如何通过pandas数据框用特定列的空格替换换行符？
I have tried this:
我试过这个：
>>> import pandas as pd
>>> lol = [[1,2,'abc','foo\nbar'], [3,1, 'def\nhaha', 'love it\n']]
>>> replace_and_strip = lambda x: x.replace('\n', ' ').strip()
>>> lol2 = [[replace_and_strip(col) if type(col) == str else col for col in list(row)] for idx, row in pd.DataFrame(lol).iterrows()]
>>> pd.DataFrame(lol2)
0 1 2 3
0 1 2 abc foo bar
1 3 1 def haha love it
But there must be a better/simpler way.
但必须有更好/更简单的方法。
回答by jezrael
回答by Wiktor Stribi?ew
You may use the following two regex replace approach:
您可以使用以下两种正则表达式替换方法：
>>> df.replace({ r'\A\s+|\s+\Z': '', '\n' : ' '}, regex=True, inplace=True)
>>> df
0 1 2 3
0 1 2 abc foo bar
1 3 1 def haha love it
>>>
Details
细节
'\A\s+|\s+\Z'->''will act likestrip()removing all leading and trailing whitespace:\A\s+- matches 1 or more whitespace symbols at the start of the string|- or\s+\Z- matches 1 or more whitespace symbols at the end of the string
'\n'->' 'will replace any newline with a space.
'\A\s+|\s+\Z'->''就像strip()删除所有前导和尾随空格：\A\s+- 匹配字符串开头的 1 个或多个空格符号|- 或者\s+\Z- 匹配字符串末尾的 1 个或多个空白符号
'\n'->' '将用空格替换任何换行符。
回答by zipa
You can select_dtypesto select columns of type objectand use applymapon those columns.
您可以select_dtypes选择类型的列object并applymap在这些列上使用。
Because there is no inplaceargument for these functions, this would be a workaround to make change to the dataframe:
因为inplace这些函数没有参数，所以这将是对数据框进行更改的一种解决方法：
strs = lol.select_dtypes(include=['object']).applymap(lambda x: x.replace('\n', ' ').strip())
lol[strs.columns] = strs
lol
# 0 1 2 3
#0 1 2 abc foo bar
#1 3 1 def haha love it
回答by Mohamed Ali JAMAOUI
Adding to the other nice answers, this is a vectorized version of your initial idea:
除了其他不错的答案之外，这是您最初想法的矢量化版本：
columns = [2,3]
df.iloc[:, columns] = [df.iloc[:,col].str.strip().str.replace('\n',' ')
for col in columns]
Details:
细节：
In [49]: df.iloc[:, columns] = [df.iloc[:,col].str.strip().str.replace('\n',' ')
for col in columns]
In [50]: df
Out[50]:
0 1 2 3
0 1 2 abc def haha
1 3 1 foo bar love it
