Python 如何为每个列名添加后缀（或前缀）？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/34049618/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to add a suffix (or prefix) to each column name?
提问by Klausos Klausos
I want to add _xsuffix to each column name like so:
我想为_x每个列名添加后缀，如下所示：
featuresA = myPandasDataFrame.columns.values + '_x'
How do I do this? Additionally, if I wanted to add x_as a suffix, how would the solution change?
我该怎么做呢？另外，如果我想添加x_作为后缀，解决方案将如何改变？
采纳答案by Stefan
You can use a listcomprehension:
你可以使用一个list理解：
df.columns = [str(col) + '_x' for col in df.columns]
There are also built-in methods like .add_suffix()and .add_prefix()as mentioned in another answer.
也有内置的类似的方法.add_suffix()，并.add_prefix()在另一个答复中提到。
回答by Jaros?aw Szymczak
The following is the nicest way to add suffix in my opinion.
以下是我认为添加后缀的最佳方式。
df = df.add_suffix('_some_suffix')
As it is a function that is called on DataFrame and returns DataFrame - you can use it in chain of the calls.
因为它是一个在 DataFrame 上调用并返回 DataFrame 的函数 - 您可以在调用链中使用它。
回答by cs95
Elegant In-place Concatenation
优雅的就地串联
If you're trying to modify dfin-place, then the cheapest (and simplest) option is in-place addition directly on df.columns(i.e., using Index.__iadd__).
如果您尝试df就地修改，那么最便宜（也是最简单）的选项是直接在df.columns（即，使用Index.__iadd__）就地添加。
df = pd.DataFrame({"A": [9, 4, 2, 1], "B": [12, 7, 5, 4]})
df
A B
0 9 12
1 4 7
2 2 5
3 1 4
df.columns += '_some_suffix'
df
A_some_suffix B_some_suffix
0 9 12
1 4 7
2 2 5
3 1 4
To add a prefix, you would similarly use
要添加前缀，您可以类似地使用
df.columns = 'some_prefix_' + df.columns
df
some_prefix_A some_prefix_B
0 9 12
1 4 7
2 2 5
3 1 4
Another cheap option is using a list comprehension with f-stringformatting (available on python3.6+).
另一个便宜的选择是使用带f-string格式的列表理解（在 python3.6+ 上可用）。
df.columns = [f'{c}_some_suffix' for c in df]
df
A_some_suffix B_some_suffix
0 9 12
1 4 7
2 2 5
3 1 4
And for prefix, similarly,
对于前缀，类似地，
df.columns = [f'some_prefix{c}' for c in df]
Method Chaining
方法链
It is also possible to do add *fixes while method chaining. To add a suffix, use DataFrame.add_suffix
也可以在方法链接时添加 *fixes。要添加后缀，请使用DataFrame.add_suffix
df.add_suffix('_some_suffix')
A_some_suffix B_some_suffix
0 9 12
1 4 7
2 2 5
3 1 4
This returns a copyof the data. IOW, dfis not modified.
这将返回数据的副本。IOW，df没有修改。
Adding prefixes is also done with DataFrame.add_prefix.
添加前缀也是用DataFrame.add_prefix.
df.add_prefix('some_prefix_')
some_prefix_A some_prefix_B
0 9 12
1 4 7
2 2 5
3 1 4
Which also does not modify df.
这也没有修改df。
Critique of add_*fix
批判 add_*fix
These are good methods if you're trying to perform method chaining:
如果您尝试执行方法链接，这些是很好的方法：
df.some_method1().some_method2().add_*fix(...)
However, add_prefix(and add_suffix) creates a copy of the entiredataframe, just to modify the headers. If you believe this is wasteful, but still want to chain, you can call pipe:
但是，add_prefix（和add_suffix）创建了整个数据帧的副本，只是为了修改标题。如果您认为这很浪费，但仍想链接，则可以调用pipe：
def add_suffix(df):
df.columns += '_some_suffix'
return df
df.some_method1().some_method2().pipe(add_suffix)
回答by JPA
I haven't seen this solution proposed above so adding this to the list:
我还没有看到上面提出的这个解决方案，所以将其添加到列表中：
df.columns += '_x'
And you can easily adapt for the prefix scenario.
您可以轻松适应前缀场景。
