如何通过根据名称而不是索引选择一列列和行来切片数据框

如何通过根据名称而不是索引选择一列列和行来切片数据框

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问题描述

这是我问的问题的后续问题。在那里我学到了一个如何做这个列(见下文)和b)在R中选择行和列似乎有很大的不同,这意味着我不能对行使用相同的方法。



所以假设我有一个这样的大熊猫数据框:

  import pandas as pd 
import numpy as np

df = pd.DataFrame(np.random.randint(10,size =(6,6)),
columns = ['c'+ str i)for i in range(6)],
index = [r+ str(i)for i in range(6)])

c0 c1 c2 c3 c4 c5
r0 4 2 3 9 9 0
r1 9 0 8 1 7 5
r2 2 6 7 5 4 7
r3 6 9 9 1 3 4
r4 1 1 1 3 0 3
r5 0 8 5 8 2 9

那么我可以轻松选择行和列的名称如下所示:

  print df.loc ['r3':'r5','c1' c4'] 

哪些返回

  c1 c2 c3 c4 
r3 9 9 1 3
r4 1 1 3 0
r5 8 5 8 2

如何在R中执行此操作?给出这样一个数据框

  df<  -  data.frame(c1 = 1:6,c2 = 2:7,c3 = 3:8,c4 = 4:9,c5 = 5:10,c6 = 6:11)
rownames(df)< - c('r1','r2','r3','r4 ','r5','r6')

c1 c2 c3 c4 c5 c6
r1 1 2 3 4 5 6
r2 2 3 4 5 6 7
r3 3 4 5 6 7 8
r4 4 5 6 7 8 9
r5 5 6 7 8 9 10
r6 6 7 8 9 10 11

显然,如果我知道我所需行/列的索引,我可以简单地做:

  df [3:5,1:4] 

但是我可能会删除分析中的行/列,以便我宁愿选择名称比索引。从上面的链接我了解到,对于列,以下将工作:

 子集(df,select = c1:c4)

其中返回

  c1 c2 c3 c4 
r1 1 2 3 4
r2 2 3 4 5
r3 3 4 5 6
r4 4 5 6 7
r5 5 6 7 8
r6 6 7 8 9

但是我还可以选择一系列行在这个特殊情况下,我可以使用 grep ,但是列如何有任意的名字?



我不想使用

  df [c('r3','r4''r5'),c('c1','c2','c3','c4')] 
解决方案

您可以使用<$($)

c $ c> which() with rownames

  subset(df [which(rownames(df)=='r3'):which(rownames(df)=='r5'),],select = c1:c4) 


c1 c2 c3 c4
r3 3 4 5 6
r4 4 5 6 7
r5 5 6 7 8


This is a follow-up question of the question I asked here. There I learned a) how to do this for columns (see below) and b) that the selection of rows and columns seems to be quite differently handled in R which means that I cannot use the same approach for rows.

So suppose I have a pandas dataframe like this:

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randint(10, size=(6, 6)),
                  columns=['c' + str(i) for i in range(6)],
                  index=["r" + str(i) for i in range(6)])

    c0  c1  c2  c3  c4  c5
r0   4   2   3   9   9   0
r1   9   0   8   1   7   5
r2   2   6   7   5   4   7
r3   6   9   9   1   3   4
r4   1   1   1   3   0   3
r5   0   8   5   8   2   9

then I can easily select rows and columns by their names like this:

print df.loc['r3':'r5', 'c1':'c4']

which returns

    c1  c2  c3  c4
r3   9   9   1   3
r4   1   1   3   0
r5   8   5   8   2

How would I do this in R? Given a dataframe like this

df <- data.frame(c1=1:6, c2=2:7, c3=3:8, c4=4:9, c5=5:10, c6=6:11)
rownames(df) <- c('r1', 'r2', 'r3', 'r4', 'r5', 'r6')

   c1 c2 c3 c4 c5 c6
r1  1  2  3  4  5  6
r2  2  3  4  5  6  7
r3  3  4  5  6  7  8
r4  4  5  6  7  8  9
r5  5  6  7  8  9 10
r6  6  7  8  9 10 11

Apparently, if I know the indexes of my desired rows/columns, I can simply do:

df[3:5, 1:4]

but I might delete rows/columns throughout my analysis so that I would rather select by name than by index. From the link above I learned that for columns the following would work:

subset(df, select=c1:c4)

which returns

  c1 c2 c3 c4
r1  1  2  3  4
r2  2  3  4  5
r3  3  4  5  6
r4  4  5  6  7
r5  5  6  7  8
r6  6  7  8  9

but how could I also select a range of rows by name at the same time?

In this particular case I could of course use grep but how about columns that have arbitrary names?

And I don't want to use

df[c('r3', 'r4' 'r5'), c('c1','c2', 'c3', 'c4')]

but an actual slice.

解决方案

You can use which() with rownames:

subset(df[which(rownames(df)=='r3'):which(rownames(df)=='r5'),], select=c1:c4)


   c1 c2 c3 c4
r3  3  4  5  6
r4  4  5  6  7
r5  5  6  7  8

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08-19 01:54