Python 剥离/修剪数据帧的所有字符串
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Strip / trim all strings of a dataframe
提问by bold
Cleaning the values of a multitype data frame in python/pandas, I want to trim the strings. I am currently doing it in two instructions :
在 python/pandas 中清理多类型数据框的值，我想修剪字符串。我目前正在按照两条说明进行操作：
import pandas as pd
df = pd.DataFrame([[' a ', 10], [' c ', 5]])
df.replace('^\s+', '', regex=True, inplace=True) #front
df.replace('\s+$', '', regex=True, inplace=True) #end
df.values
This is quite slow, what could I improve ?
这很慢，我可以改进什么？
回答by jezrael
You can use DataFrame.select_dtypesto select stringcolumns and then applyfunction str.strip.
您可以使用DataFrame.select_dtypes来选择string列，然后apply使用str.strip.
Notice: Values cannot be typeslike dictsor lists, because their dtypesis object.
注意：值不能types像dicts或lists，因为它们dtypes是object。
df_obj = df.select_dtypes(['object'])
print (df_obj)
0 a
1 c
df[df_obj.columns] = df_obj.apply(lambda x: x.str.strip())
print (df)
0 1
0 a 10
1 c 5
But if there are only a few columns use str.strip:
但如果只有几列，请使用str.strip：
df[0] = df[0].str.strip()
回答by Jonathan B.
Money Shot
金钱射击
Here's a compact version of using applymapwith a straightforward lambda expression to call striponly when the value is of a string type:
这是使用applymap简单的 lambda 表达式的紧凑版本，strip仅当值是字符串类型时才调用：
df.applymap(lambda x: x.strip() if isinstance(x, str) else x)
Full Example
完整示例
A more complete example:
一个更完整的例子：
import pandas as pd
def trim_all_columns(df):
"""
Trim whitespace from ends of each value across all series in dataframe
"""
trim_strings = lambda x: x.strip() if isinstance(x, str) else x
return df.applymap(trim_strings)
# simple example of trimming whitespace from data elements
df = pd.DataFrame([[' a ', 10], [' c ', 5]])
df = trim_all_columns(df)
print(df)
>>>
0 1
0 a 10
1 c 5
Working Example
工作示例
Here's a working example hosted by trinket: https://trinket.io/python3/e6ab7fb4ab
这是由饰品托管的工作示例：https: //trinket.io/python3/e6ab7fb4ab
回答by Roman Pekar
If you really want to use regex, then
如果你真的想使用正则表达式，那么
>>> df.replace('(^\s+|\s+$)', '', regex=True, inplace=True)
>>> df
0 1
0 a 10
1 c 5
But it should be faster to do it like this:
但是这样做应该更快：
>>> df[0] = df[0].str.strip()
回答by Aakash Makwana
You can try:
你可以试试：
df[0] = df[0].str.strip()
or more specifically for all string columns
或更具体地针对所有字符串列
non_numeric_columns = list(set(df.columns)-set(df._get_numeric_data().columns))
df[non_numeric_columns] = df[non_numeric_columns].apply(lambda x : str(x).strip())
回答by Dekel
You can use the applyfunctionof the Seriesobject:
您可以使用该apply功能的的Series对象：
>>> df = pd.DataFrame([[' a ', 10], [' c ', 5]])
>>> df[0][0]
' a '
>>> df[0] = df[0].apply(lambda x: x.strip())
>>> df[0][0]
'a'
Note the usage of
stripand not theregexwhich is much faster
请注意使用
stripand not theregex哪个更快
Another option - use the applyfunctionof the DataFrame object:
另一种选择 - 使用DataFrame 对象的apply功能：
>>> df = pd.DataFrame([[' a ', 10], [' c ', 5]])
>>> df.apply(lambda x: x.apply(lambda y: y.strip() if type(y) == type('') else y), axis=0)
0 1
0 a 10
1 c 5
回答by hyunwoo jeong
def trim(x):
if x.dtype == object:
x = x.str.split(' ').str[0]
return(x)
df = df.apply(trim)
