逐行乘以数据框 [英] Multiply a data frame row-by-row
问题描述
输入文件:
df1 <- data.frame(row.names=c("w","x","y","z"),
A=c(0,0,0,0),
B=c(0,1,0,0),
C=c(1,0,1,0),
D=c(1,1,1,1))
A B C D
w 0 0 1 1
x 0 1 0 1
y 0 0 1 1
z 0 0 0 1
我想应用一个方程,即行 w 乘以行 x 以获得 w-x 对的成对值,如下所示:
I want to apply an equation i.e. multiply row w to row x to get the pairwise value for w-x pair, as follows:
A B C D
w 0 0 1 1
X x 0 1 0 1
--------------
wx 0 0 0 1
对 w-x、w-y、w-y、w-z、x-y、x-z、y-z 进行逐行分析.并生成一个具有 6 列的新数据框(两个行名称,后跟相乘值).
to get row-wise analysis for w-x, w-y, w-y, w-z, x-y, x-z, y-z. and generate a new dataframe with 6 columns (two row names followed by the multiplied values).
就是这样
w x 0 0 0 1
w y 0 0 1 1
w z 0 0 0 1
x y 0 0 0 1
x z 0 0 0 1
y z 0 0 0 1
谢谢.
推荐答案
如果您不想在结果对象中使用组合名称,那么我们可以结合@DWin's 和@Owen's Answers 的元素来提供真正的矢量化方法问题.(您可以将组合名称添加为行名称,并在末尾添加一个额外步骤.)
If you don't want the combo names in the resulting object, then we can combine elements of @DWin's and @Owen's Answers to provide a truly vectorised approach to the problem. (You can add the combination names as row names with one extra step at the end.)
一、数据:
dat <- read.table(con <- textConnection(" A B C D
w 0 0 1 1
x 0 1 0 1
y 0 0 1 1
z 0 0 0 1
"), header=TRUE)
close(con)
从@DWin 的答案中获取 combn()
的想法,但在 dat
的行索引上使用它:
Take the combn()
idea from @DWin's Answer but use it on the row indices of dat
:
combs <- combn(seq_len(nrow(dat)), 2)
combs
的行现在索引我们想要相乘的 dat
的行:
The rows of combs
now index the rows of dat
that we want to multiply together:
> combs
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 1 2 2 3
[2,] 2 3 4 3 4 4
现在我们采用@Owen 展示的想法,即 dat[i, ] * dat[j, ]
和 i
和 j
combs
的第一行和第二行.我们使用 data.matrix()
转换为矩阵,因为这对于大型对象会更有效,但代码也将使用 dat
作为数据框.>
Now we take the idea @Owen showed, namely dat[i, ] * dat[j, ]
with i
and j
being the first and second rows of combs
respectively. We convert to a matrix with data.matrix()
as this will be more efficient for large objects, but the code will work with dat
as a data frame too.
mat <- data.matrix(dat)
mat[combs[1,], ] * mat[combs[2,], ]
产生:
> mat[combs[1,], ] * mat[combs[2,], ]
A B C D
w 0 0 0 1
w 0 0 1 1
w 0 0 0 1
x 0 0 0 1
x 0 0 0 1
y 0 0 0 1
要了解其工作原理,请注意 mat[combs[k,], ]
生成一个矩阵,其中各行按组合指定的顺序重复:
To see how this works, note that mat[combs[k,], ]
produces a matrix with various rows repeated in the order specified by the combinations:
> mat[combs[1,], ]
A B C D
w 0 0 1 1
w 0 0 1 1
w 0 0 1 1
x 0 1 0 1
x 0 1 0 1
y 0 0 1 1
> mat[combs[2,], ]
A B C D
x 0 1 0 1
y 0 0 1 1
z 0 0 0 1
y 0 0 1 1
z 0 0 0 1
z 0 0 0 1
为了准确获取 OP 发布的内容,我们可以使用第二个 combn()
调用修改行名:
To get exactly what the OP posted, we can modify the rownames using a second combn()
call:
> out <- mat[combs[1,], ] * mat[combs[2,], ]
> rownames(out) <- apply(combn(rownames(dat), 2), 2, paste, collapse = "")
> out
A B C D
wx 0 0 0 1
wy 0 0 1 1
wz 0 0 0 1
xy 0 0 0 1
xz 0 0 0 1
yz 0 0 0 1
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