Python 删除熊猫数据框中的特殊字符
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Remove special characters in pandas dataframe
提问by RageQuilt
This seems like an inherently simple task but I am finding it very difficult to remove the '' from my entire data frame and return the numeric values in each column, including the numbers that did not have ''. The dateframe includes hundreds of more columns and looks like this in short:
这似乎是一项本质上很简单的任务，但我发现很难从整个数据框中删除 ' ' 并返回每列中的数值，包括没有 ''的数字。日期框架包含数百个列，简而言之如下所示：
Time A1 A2
2.0002546296 1499 1592
2.0006712963 1252 1459
2.0902546296 1731 2223
2.0906828704 1691 1904
2.1742245370 2364 3121
2.1764699074 2096 1942
2.7654050926 *7639* *8196*
2.7658564815 *7088* *7542*
2.9048958333 *8736* *8459*
2.9053125000 *7778* *7704*
2.9807175926 *6612* *6593*
3.0585763889 *8520* *9122*
I have not written it to iterate over every column in df yet but as far as the first column goes I have come up with this
我还没有写它来迭代 df 中的每一列，但就第一列而言，我想出了这个
df['A1'].str.replace('*','').astype(float)
which yields
这产生
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
5 NaN
6 NaN
7 NaN
8 NaN
9 NaN
10 NaN
11 NaN
12 NaN
13 NaN
14 NaN
15 NaN
16 NaN
17 NaN
18 NaN
19 7639.0
20 7088.0
21 8736.0
22 7778.0
23 6612.0
24 8520.0
Is there a very easy way to just remove the '*' in the dataframe in pandas?
有没有一种非常简单的方法可以删除熊猫数据框中的“*”？
回答by shivsn
use replacewhich applies on whole dataframe :
使用适用于整个数据框的替换：
df
Out[14]:
Time A1 A2
0 2.000255 1499 1592
1 2.176470 2096 1942
2 2.765405 *7639* *8196*
3 2.765856 *7088* *7542*
4 2.904896 *8736* *8459*
5 2.905312 *7778* *7704*
6 2.980718 *6612* *6593*
7 3.058576 *8520* *9122*
df=df.replace('\*','',regex=True).astype(float)
df
Out[16]:
Time A1 A2
0 2.000255 1499 1592
1 2.176470 2096 1942
2 2.765405 7639 8196
3 2.765856 7088 7542
4 2.904896 8736 8459
5 2.905312 7778 7704
6 2.980718 6612 6593
7 3.058576 8520 9122
回答by amin
There is another solution which uses map and strip functions. You can see the below link: Pandas DataFrame: remove unwanted parts from strings in a column.
还有另一种使用 map 和 strip 函数的解决方案。您可以看到以下链接： Pandas DataFrame：从列中的字符串中删除不需要的部分。
df =
Time A1 A2
0 2.0 1258 *1364*
1 2.1 *1254* 2002
2 2.2 1520 3364
3 2.3 *300* *10056*
cols = ['A1', 'A2']
for col in cols:
df[col] = df[col].map(lambda x: str(x).lstrip('*').rstrip('*')).astype(float)
df =
Time A1 A2
0 2.0 1258 1364
1 2.1 1254 2002
2 2.2 1520 3364
3 2.3 300 10056
The parsing procedure only be applied on the desired columns.
解析过程仅应用于所需的列。
回答by CuriousCoder
I found this to be a simple approach - Use replaceto retain only the digits (and dotand minussign).
This would remove characters, alphabets or anything that is not defined in to_replaceattribute.
我发现这是一种简单的方法 - 用于replace仅保留数字（dot和minus符号）。
这将删除字符、字母或任何未在to_replace属性中定义的内容。
So, the solution is:df['A1'].replace(regex=True, inplace=True, to_replace=r'[^0-9.\-]', value=r'']df['A1'] = df['A1'].astype(float64)
所以，解决办法是：df['A1'].replace(regex=True, inplace=True, to_replace=r'[^0-9.\-]', value=r'']df['A1'] = df['A1'].astype(float64)
回答by ?oàn Ph??ng Th?o
I found the answer of CuriousCoder so brief and useful but there must be a ')'instead of ']'So it should be:
我发现 CuriousCoder 的答案如此简短和有用，但必须有一个')'而不是']'所以它应该是：
df['A1'].replace(regex=True, inplace=True, to_replace=r'[^0-9.\-]',
value=r''] df['A1'] = df['A1'].astype(float64)
