将值设置为熊猫数据框的整个列

我试图将一个数据框架的整个列设置为一个特定的值。

In  [1]: df
Out [1]:
issueid   industry
0        001        xxx
1        002        xxx
2        003        xxx
3        004        xxx
4        005        xxx

在我看来，loc是在数据框架中替换值的最佳实践(不是吗?) :

In  [2]: df.loc[:,'industry'] = 'yyy'

然而，我仍然收到了这样一条经常被谈论的警告信息:

A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead

如果我愿意的话

In  [3]: df['industry'] = 'yyy'

我收到了同样的警告信息。

有什么想法吗? 使用 Python3.5.2和熊猫0.18.1。

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小开

You can do : 

df['industry'] = 'yyy'
小开

Assuming your Data frame is like 'Data' you have to consider if your data is a string or an integer. Both are treated differently. So in this case you need be specific about that.

import pandas as pd


data = [('001','xxx'), ('002','xxx'), ('003','xxx'), ('004','xxx'), ('005','xxx')]


df = pd.DataFrame(data,columns=['issueid', 'industry'])


print("Old DataFrame")
print(df)


df.loc[:,'industry'] = str('yyy')


print("New DataFrame")
print(df)

Now if want to put numbers instead of letters you must create and array

list_of_ones = [1,1,1,1,1]
df.loc[:,'industry'] = list_of_ones
print(df)

Or if you are using Numpy

import numpy as np
n = len(df)
df.loc[:,'industry'] = np.ones(n)
print(df)
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最佳答案

Python can do unexpected things when new objects are defined from existing ones. You stated in a comment above that your dataframe is defined along the lines of df = df_all.loc[df_all['issueid']==specific_id,:]. In this case, df is really just a stand-in for the rows stored in the df_all object: a new object is NOT created in memory.

To avoid these issues altogether, I often have to remind myself to use the copy module, which explicitly forces objects to be copied in memory so that methods called on the new objects are not applied to the source object. I had the same problem as you, and avoided it using the deepcopy function.

In your case, this should get rid of the warning message:

from copy import deepcopy
df = deepcopy(df_all.loc[df_all['issueid']==specific_id,:])
df['industry'] = 'yyy'

EDIT: Also see David M.'s excellent comment below!

df = df_all.loc[df_all['issueid']==specific_id,:].copy()
df['industry'] = 'yyy'
小开

This provides you with the possibility of adding conditions on the rows and then change all the cells of a specific column corresponding to those rows:

df.loc[(df['issueid'] == '001'), 'industry'] = str('yyy')
小开

You can use the assign function:

df = df.assign(industry='yyy')
小开 
df.loc[:,'industry'] = 'yyy'

This does the magic. You are to add '.loc' with ':' for all rows. Hope it helps

小开

Seems to me that:

df1 = df[df['col1']==some_value] will not create a new DataFrame, basically, changes in df1 will be reflected in the parent df. This leads to the warning.
Whereas, df1 = df[df['col1]]==some_value].copy() will create a new DataFrame, and changes in df1 will not be reflected in df. The copy method is recommended if you don't want to make changes to your original df.

小开

I had a similar issue before even with this approach df.loc[:,'industry'] = 'yyy', but once I refreshed the notebook, it ran well.

You may want to try refreshing the cells after you have df.loc[:,'industry'] = 'yyy'.

小开

if you just create new but empty data frame, you cannot directly sign a value to a whole column. This will show as NaN because the system wouldn't know how many rows the data frame will have!You need to either define the size or have some existing columns.

df = pd.DataFrame()
df["A"] = 1
df["B"] = 2
df["C"] = 3
小开

Only use them instead:

df.iloc[:]['industry'] = 'yyy'

remember: this only works with exist columns in dataframe

this for people who didn't work .loc


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