计算数据帧中每对列之间的成对差异 [英] Calculate pairwise-difference between each pair of columns in dataframe

问题描述

我无法解决创建adat中每个变量（列）的差异的问题，并将其保存到矩阵dfmtx。

I cannot get around the problem of creating the differentials of every variable (column) in "adat" and saving it to a matrix "dfmtx".

我会只需要自动化以下序列来运行adat中的每一列，而不是按照相应名称的名字命名所获取的矢量，并将其放在dfmtx列中。

I would just need to automate the following sequence to run for each column in "adat" and than name the obtained vector according to the name of the ones subtracted from each other and placed in to a column of "dfmtx".

在adat中，我有14列和26行不包括标题。

In "adat" I have 14 columns and 26 rows not including the header.

dfmtx[,1]=(adat[,1]-adat[,1])
dfmtx[,2]=(adat[,1]-adat[,2])
dfmtx[,3]=(adat[,1]-adat[,3])
dfmtx[,4]=(adat[,1]-adat[,4])
dfmtx[,5]=(adat[,1]-adat[,5])
dfmtx[,6]=(adat[,1]-adat[,6])
.....
dfmtx[,98]=(adat[,14]-adat[,14])

任何帮助将不胜感激

推荐答案

如果 adat 是一个data.frame，你可以使用外部获取列的组合，然后进行差异下注根据 outer 的索引，将列的成对子集删除。不清楚你如何获得98栏。通过删除对角线和下三角形元素，列数将为91。

If adat is a data.frame, you can use outer to get the combinations of columns and then do the difference between pairwise subset of columns based on the index from outer. It is not clear how you got "98" columns. By removing the diagonal and lower triangular elements, the number of columns will be "91".

nm1 <- outer(colnames(adat), colnames(adat), paste, sep="_")
indx1 <-  which(lower.tri(nm1, diag=TRUE))
res <- outer(1:ncol(adat), 1:ncol(adat), 
              function(x,y) adat[,x]-adat[,y])
colnames(res) <- nm1
res1 <- res[-indx1]
dim(res1)
#[1] 26 91

数据

data

set.seed(24)
adat <- as.data.frame(matrix(sample(1:20, 26*14,
                         replace=TRUE), ncol=14))

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