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2013-07-10
如果有下列矩阵,如何根据每一行的最大值找到对应列名呢?请大侠帮忙!
             A                        B                C                      D
1  3.773964e-05 2.236609e-07 1.204275e-06 9.999608e-01
2  7.805035e-07 1.026306e-10 1.507551e-07 9.999991e-01
3  5.553381e-06 2.985323e-08 1.000433e-06 9.999934e-01
4  7.893392e-07 9.274931e-11 1.441299e-07 9.999991e-01
5  3.000040e-04 2.510037e-02 2.350496e-04 9.743646e-01


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2013-7-10 12:30:57
x <- matrix(c(3.773964e-05, 2.236609e-07, 1.204275e-06, 9.999608e-01,
              7.805035e-07, 1.026306e-10, 1.507551e-07, 9.999991e-01,
              5.553381e-06, 2.985323e-08, 1.000433e-06, 9.999934e-01,
              7.893392e-07, 9.274931e-11, 1.441299e-07, 9.999991e-01,
              3.000040e-04, 2.510037e-02, 2.350496e-04, 9.743646e-01),
             nrow = 5, ncol = 4, byrow = TRUE)
dimnames(x) <- list(1:5, LETTERS[1:4])
apply(x, 1, function(t) colnames(x)[which.max(t)])
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2013-7-10 15:07:00
hugebear 发表于 2013-7-10 12:30
x
谢谢你,问题解决!
学习下apply()函数
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2013-7-12 11:04:20
> x <- matrix(c(3.773964e-05, 2.236609e-07, 1.204275e-06, 9.999608e-01,
+               7.805035e-07, 1.026306e-10, 1.507551e-07, 9.999991e-01,
+               5.553381e-06, 2.985323e-08, 1.000433e-06, 9.999934e-01,
+               7.893392e-07, 9.274931e-11, 1.441299e-07, 9.999991e-01,
+               3.000040e-04, 2.510037e-02, 2.350496e-04, 9.743646e-01),
+             nrow = 5, ncol = 4, byrow = TRUE)
> x
             [,1]         [,2]         [,3]      [,4]
[1,] 3.773964e-05 2.236609e-07 1.204275e-06 0.9999608
[2,] 7.805035e-07 1.026306e-10 1.507551e-07 0.9999991
[3,] 5.553381e-06 2.985323e-08 1.000433e-06 0.9999934
[4,] 7.893392e-07 9.274931e-11 1.441299e-07 0.9999991
[5,] 3.000040e-04 2.510037e-02 2.350496e-04 0.9743646
> dimnames(x) <- list(1:5, LETTERS[1:4])
> x
             A            B            C         D
1 3.773964e-05 2.236609e-07 1.204275e-06 0.9999608
2 7.805035e-07 1.026306e-10 1.507551e-07 0.9999991
3 5.553381e-06 2.985323e-08 1.000433e-06 0.9999934
4 7.893392e-07 9.274931e-11 1.441299e-07 0.9999991
5 3.000040e-04 2.510037e-02 2.350496e-04 0.9743646
> apply(x, 1, function(t) colnames(x)[which.max(t)])
  1   2   3   4   5
"D" "D" "D" "D" "D"
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2013-7-12 11:05:45
apply {base}        R Documentation
Apply Functions Over Array Margins

Description

Returns a vector or array or list of values obtained by applying a function to margins of an array or matrix.

Usage

apply(X, MARGIN, FUN, ...)
Arguments

X       
an array, including a matrix.

MARGIN       
a vector giving the subscripts which the function will be applied over. E.g., for a matrix 1 indicates rows, 2 indicates columns, c(1, 2) indicates rows and columns. Where X has named dimnames, it can be a character vector selecting dimension names.

FUN       
the function to be applied: see ‘Details’. In the case of functions like +, %*%, etc., the function name must be backquoted or quoted.

...       
optional arguments to FUN.

Details

If X is not an array but an object of a class with a non-null dim value (such as a data frame), apply attempts to coerce it to an array via as.matrix if it is two-dimensional (e.g., a data frame) or via as.array.

FUN is found by a call to match.fun and typically is either a function or a symbol (e.g. a backquoted name) or a character string specifying a function to be searched for from the environment of the call to apply.

Value

If each call to FUN returns a vector of length n, then apply returns an array of dimension c(n, dim(X)[MARGIN]) if n > 1. If n equals 1, apply returns a vector if MARGIN has length 1 and an array of dimension dim(X)[MARGIN] otherwise. If n is 0, the result has length 0 but not necessarily the ‘correct’ dimension.

If the calls to FUN return vectors of different lengths, apply returns a list of length prod(dim(X)[MARGIN]) with dim set to MARGIN if this has length greater than one.

In all cases the result is coerced by as.vector to one of the basic vector types before the dimensions are set, so that (for example) factor results will be coerced to a character array.

References

Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.

See Also

lapply and there, simplify2array; tapply, and convenience functions sweep and aggregate.

Examples

## Compute row and column sums for a matrix:
x <- cbind(x1 = 3, x2 = c(4:1, 2:5))
dimnames(x)[[1]] <- letters[1:8]
apply(x, 2, mean, trim = .2)
col.sums <- apply(x, 2, sum)
row.sums <- apply(x, 1, sum)
rbind(cbind(x, Rtot = row.sums), Ctot = c(col.sums, sum(col.sums)))

stopifnot( apply(x, 2, is.vector))

## Sort the columns of a matrix
apply(x, 2, sort)

##- function with extra args:
cave <- function(x, c1, c2) c(mean(x[c1]), mean(x[c2]))
apply(x,1, cave,  c1="x1", c2=c("x1","x2"))

ma <- matrix(c(1:4, 1, 6:8), nrow = 2)
ma
apply(ma, 1, table)  #--> a list of length 2
apply(ma, 1, stats::quantile)# 5 x n matrix with rownames

stopifnot(dim(ma) == dim(apply(ma, 1:2, sum)))

## Example with different lengths for each call
z <- array(1:24, dim=2:4)
zseq <- apply(z, 1:2, function(x) seq_len(max(x)))
zseq         ## a 2 x 3 matrix
typeof(zseq) ## list
dim(zseq) ## 2 3
zseq[1,]
apply(z, 3, function(x) seq_len(max(x)))
# a list without a dim attribute
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2018-10-12 10:10:17
hugebear 发表于 2013-7-10 12:30
x
请问一下,我现在已知矩阵中的某个元素,想把该元素所在列的列名表示出来,看列名是否在数组(A,B,C,D,E,F,G)中,请问代码应该怎么写?谢谢
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