使用pandas数据框中的一行而无需进行链索引编制(不应对仅索引编制) [英] Work with a row in a pandas dataframe without incurring chain indexing (not coping just indexing)

问题描述

我的数据组织在一个数据框中:

My data is organized in a dataframe:

import pandas as pd
import numpy as np

data = {'Col1' : [4,5,6,7], 'Col2' : [10,20,30,40], 'Col3' : [100,50,-30,-50], 'Col4' : ['AAA', 'BBB', 'AAA', 'CCC']}

df = pd.DataFrame(data=data, index = ['R1','R2','R3','R4'])

看起来像这样(只有更大):

Which looks like this (only much bigger):

    Col1  Col2  Col3 Col4
R1     4    10   100  AAA
R2     5    20    50  BBB
R3     6    30   -30  AAA
R4     7    40   -50  CCC

我的算法遍历此表行并执行一组操作.

My algorithm loops through this table rows and performs a set of operations.

出于清洁/懒惰的考虑，我希望在每次迭代时只处理一行，而无需键入df.loc['row index', 'column name']来获取每个单元格值

For cleaness/lazyness sake, I would like to work on a single row at each iteration without typing df.loc['row index', 'column name'] to get each cell value

我尝试遵循正确的样式例如:

row_of_interest = df.loc['R2', :]

但是，我仍然会收到警告:

However, I still get the warning when I do:

row_of_interest['Col2'] = row_of_interest['Col2'] + 1000

SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame

它无法正常工作(按我的预期)，正在制作副本

And it is not working (as I intended) it is making a copy

print df

    Col1  Col2  Col3 Col4
R1     4    10   100  AAA
R2     5    20    50  BBB
R3     6    30   -30  AAA
R4     7    40   -50  CCC

关于正确方法的任何建议吗?还是我应该坚持直接使用数据框?

Any advice on the proper way to do it? Or should I just stick to work with the data frame directly?

使用提供的答复，从代码中删除了警告，但未修改原始数据框:感兴趣的行" Series是副本，并非原始数据框的一部分.例如:

Using the replies provided the warning is removed from the code but the original dataframe is not modified: The "row of interest" Series is a copy not part of the original dataframe. For example:

import pandas as pd
import numpy as np

data = {'Col1' : [4,5,6,7], 'Col2' : [10,20,30,40], 'Col3' : [100,50,-30,-50], 'Col4' : ['AAA', 'BBB', 'AAA', 'CCC']}

df = pd.DataFrame(data=data, index = ['R1','R2','R3','R4'])

row_of_interest         = df.loc['R2']
row_of_interest.is_copy = False
new_cell_value          = row_of_interest['Col2'] + 1000
row_of_interest['Col2'] = new_cell_value

print row_of_interest 

Col1       5
Col2    1020
Col3      50
Col4     BBB
Name: R2, dtype: object

print df

    Col1  Col2  Col3 Col4
R1     4    10   100  AAA
R2     5    20    50  BBB
R3     6    30   -30  AAA
R4     7    40   -50  CCC

这是我要复制的功能的示例.在python中，列表列表如下所示:

This is an example of the functionality I would like to replicate. In python a list of lists looks like:

a = [[1,2,3],[4,5,6]]

现在我可以创建一个标签"

Now I can create a "label"

b = a[0]

如果我更改b中的条目:

And if I change an entry in b:

b[0] = 7

a和b都改变了.

print a, b

[[7,2,3],[4,5,6]], [7,2,3]

可以在标有熊猫系列之一的熊猫数据框之间复制这种行为吗?

Can this behavior be replicated between a pandas dataframe labeling one of its rows a pandas series?

推荐答案

这应该有效:

row_of_interest = df.loc['R2', :]
row_of_interest.is_copy = False
row_of_interest['Col2'] = row_of_interest['Col2'] + 1000

设置.is_copy = False是诀窍

import pandas as pd
import numpy as np

data = {'Col1' : [4,5,6,7], 'Col2' : [10,20,30,40], 'Col3' : [100,50,-30,-50], 'Col4' : ['AAA', 'BBB', 'AAA', 'CCC']}

df = pd.DataFrame(data=data, index = ['R1','R2','R3','R4'])

row_of_interest         = df.loc['R2']
row_of_interest.is_copy = False
new_cell_value          = row_of_interest['Col2'] + 1000
row_of_interest['Col2'] = new_cell_value

print row_of_interest 

df.loc['R2'] = row_of_interest 

print df

df:

    Col1  Col2  Col3 Col4
R1     4    10   100  AAA
R2     5  1020    50  BBB
R3     6    30   -30  AAA
R4     7    40   -50  CCC

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