[英]sorting data frame columns based on specific value in each column
I am using the Tidyverse package in R. I have a data frame with 20 rows and 500 columns. 我在R中使用Tidyverse包。我有一个包含20行500列的数据框。 I want to sort all the columns based on the size of the value in the last row of each column.
我想根据每列最后一行中值的大小对所有列进行排序。
Here is an example with just 3 rows and 4 columns: 这是一个只有3行4列的示例:
1 2 3 4,
5 6 7 8,
8 7 9 1
The desired result is: 理想的结果是:
3 1 2 4,
7 5 6 8,
9 8 7 1
I searched stack overflow but could not find an answer to this type of question. 我搜索了堆栈溢出,但找不到此类问题的答案。
The following reorders the data frame columns by the order of the last-rows values: 以下内容按最后一行值的顺序对数据框列进行重新排序:
df <- data.frame(col1=c(1,5,8),col2=c(2,6,7),col3=c(3,7,9),col4=c(4,8,1))
last_row <- df[nrow(df),]
df <- df[,order(last_row,decreasing = T)]
First, to get the last rows. 首先,获取最后一行。 Then to sort them with the order() function and to return the reordered columns.
然后使用order()函数对它们进行排序,并返回重新排序的列。
>df
col3 col1 col2 col4
1 3 1 2 4
2 7 5 6 8
3 9 8 7 1
If we want to use dplyr
from tidyverse
, we can use slice
to get the last row and then use order
in decreasing
order to subset columns. 如果要使用
dplyr
的tidyverse
,可以使用slice
来获取最后一行,然后order
decreasing
使用order来对列进行子集化。
library(dplyr)
df[df %>% slice(n()) %>% order(decreasing = TRUE)]
# V3 V1 V2 V4
#1 3 1 2 4
#2 7 5 6 8
#3 9 8 7 1
Whose translation in base R would be R基为谁的翻译是
df[order(df[nrow(df), ], decreasing = TRUE)]
data 数据
df <- read.table(text = "1 2 3 4
5 6 7 8
8 7 9 1")
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