Python 将列中的所有值复制到 Pandas 数据框中的新列
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Copy all values in a column to a new column in a pandas dataframe
提问by Justin Buchanan
This is a very basic question, I just can not seem to find an answer.
这是一个非常基本的问题，我似乎无法找到答案。
I have a dataframe like this, called df:
我有一个这样的数据框，叫做 df：
A B C
a.1 b.1 c.1
a.2 b.2 c.2
a.3 b.3 c.3
Then I extract all the rows from df, where column 'B' has a value of 'b.2'. I assign these results to df_2.
然后我从 df 中提取所有行，其中列 'B' 的值为 'b.2'。我将这些结果分配给 df_2。
df_2 = df[df['B'] == 'b.2']
df_2 becomes:
df_2 变为：
A B C
a.2 b.2 c.2
Then, I copy all the values in column 'B' to a new column named 'D'. Causing df_2 to become:
然后，我将“B”列中的所有值复制到名为“D”的新列中。导致 df_2 变为：
A B C D
a.2 b.2 c.2 b.2
When I preform an assignment like this:
当我执行这样的作业时：
df_2['D'] = df_2['B']
I get the following warning:
我收到以下警告：
A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead
See the the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
试图在来自 DataFrame 的切片副本上设置值。尝试使用 .loc[row_indexer,col_indexer] = value 代替
请参阅文档中的警告：http: //pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
I have also tried using .loc when creating df_2 like this:
在创建 df_2 时，我也尝试使用 .loc ，如下所示：
df_2 = df.loc[df['B'] == 'b.2']
However, I still get the warning.
但是，我仍然收到警告。
Any help is greatly appreciated.
任何帮助是极大的赞赏。
采纳答案by Anand S Kumar
You can simply assign the Bto the new column , Like -
您可以简单地将 分配B给新列，例如 -
df['D'] = df['B']
Example/Demo -
示例/演示 -
In [1]: import pandas as pd
In [2]: df = pd.DataFrame([['a.1','b.1','c.1'],['a.2','b.2','c.2'],['a.3','b.3','c.3']],columns=['A','B','C'])
In [3]: df
Out[3]:
A B C
0 a.1 b.1 c.1
1 a.2 b.2 c.2
2 a.3 b.3 c.3
In [4]: df['D'] = df['B'] #<---What you want.
In [5]: df
Out[5]:
A B C D
0 a.1 b.1 c.1 b.1
1 a.2 b.2 c.2 b.2
2 a.3 b.3 c.3 b.3
In [6]: df.loc[0,'D'] = 'd.1'
In [7]: df
Out[7]:
A B C D
0 a.1 b.1 c.1 d.1
1 a.2 b.2 c.2 b.2
2 a.3 b.3 c.3 b.3
回答by Alex
The problem is in the line before the one that throws the warning. When you create df_2 that's where you're creating a copy of a slice of a dataframe. Instead, when you create df_2, use .copy() and you won't get that warning later on.
问题出在发出警告的那一行之前。当您创建 df_2 时，您正在创建数据帧切片的副本。相反，当您创建 df_2 时，使用 .copy() 并且您以后不会收到该警告。
df_2 = df[df['B'] == 'b.2'].copy()
回答by Luke
I think the correct access method is using the index:
我认为正确的访问方法是使用索引：
df_2.loc[:,'D'] = df_2['B']
回答by Mark Andersen
Following up on these solutions, here is some helpful code illustrating :
跟进这些解决方案，这里有一些有用的代码说明：
#
# Copying columns in pandas without slice warning
#
import numpy as np
df = pd.DataFrame(np.random.randn(10, 3), columns=list('ABC'))
#
# copies column B into new column D
df.loc[:,'D'] = df['B']
print df
#
# creates new column 'E' with values -99
#
# But copy command replaces those where 'B'>0 while others become NaN (not copied)
df['E'] = -99
print df
df['E'] = df[df['B']>0]['B'].copy()
print df
#
# creates new column 'F' with values -99
#
# Copy command only overwrites values which meet criteria 'B'>0
df['F']=-99
df.loc[df['B']>0,'F'] = df[df['B']>0]['B'].copy()
print df
回答by KarthikS
How about:
怎么样：
df['D'] = df['B'].values
df['D'] = df['B'].values
