R-table和tapply函数

table可统计数据的频数

tapply可根据因子、向量和要计算的函数计算

> class<-c(1,2,3,2,1,2,1,3)

> class
[1] 1 2 3

> c(81,65,72,88,73,91,56,90)->student
> class
[1] 1 2 3 2 1 2 1 3

>factor(class)->class

> tapply(student,class,mean)
1 2 3
70.00000 81.33333 81.00000
> tapply(student,class,min)
1 2 3
56 65 72

> tapply(student,class,max)
1 2 3
81 91 90
> table(class)
class
1 2 3
3 3 2
>

Apply a Function Over a Ragged Array

Description

Apply a function to each cell of a ragged array, that is to each (non-empty) group of values given by a unique combination of the levels of certain factors.

Usage

tapply(X, INDEX, FUN = NULL, ..., simplify = TRUE)

Arguments

`X`	an atomic object, typically a vector.
`INDEX`	list of factors, each of same length as `X`. The elements are coerced to factors by `as.factor`.
`FUN`	the function to be applied, or `NULL`. In the case of functions like `+`, `%*%`, etc., the function name must be backquoted or quoted. If `FUN` is `NULL`, tapply returns a vector which can be used to subscript the multi-way array `tapply` normally produces.
`...`	optional arguments to `FUN`: the Note section.
`simplify`	If `FALSE`, `tapply` always returns an array of mode `"list"`. If `TRUE` (the default), then if `FUN` always returns a scalar, `tapply` returns an array with the mode of the scalar.

Value

If FUN is not NULL, it is passed to match.fun, and hence it can be a function or a symbol or character string naming a function.

When FUN is present, tapply calls FUN for each cell that has any data in it. If FUN returns a single atomic value for each such cell (e.g., functions mean or var) and when simplify is TRUE,tapply returns a multi-way array containing the values, and NA for the empty cells. The array has the same number of dimensions as INDEX has components; the number of levels in a dimension is the number of levels (nlevels()) in the corresponding component of INDEX. Note that if the return value has a class (e.g. an object of class "Date") the class is discarded.

Note that contrary to S, simplify = TRUE always returns an array, possibly 1-dimensional.

If FUN does not return a single atomic value, tapply returns an array of mode list whose components are the values of the individual calls to FUN, i.e., the result is a list with a dimattribute.

When there is an array answer, its dimnames are named by the names of INDEX and are based on the levels of the grouping factors (possibly after coercion).

For a list result, the elements corresponding to empty cells are NULL.

Note

Optional arguments to FUN supplied by the ... argument are not divided into cells. It is therefore inappropriate for FUN to expect additional arguments with the same length as X.

References

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

Examples

require(stats) groups <- as.factor(rbinom(32, n = 5, prob = 0.4)) tapply(groups, groups, length) #- is almost the same as table(groups) ## contingency table from data.frame : array with named dimnames tapply(warpbreaks$breaks, warpbreaks[,-1], sum) tapply(warpbreaks$breaks, warpbreaks[, 3, drop = FALSE], sum) n <- 17; fac <- factor(rep(1:3, length = n), levels = 1:5) table(fac) tapply(1:n, fac, sum) tapply(1:n, fac, sum, simplify = FALSE) tapply(1:n, fac, range) tapply(1:n, fac, quantile) ## example of ... argument: find quarterly means tapply(presidents, cycle(presidents), mean, na.rm = TRUE) ind <- list(c(1, 2, 2), c("A", "A", "B")) table(ind) tapply(1:3, ind) #-> the split vector tapply(1:3, ind, sum)

问题：

有数万个数据,两列数据一列为名称(A列 ) 一列为值（x列），一个名称可对应多个值，一个值可能有多个名称，具体问题如下所示
A1 x1
A2 x2
A3 x3
A4 x4
A1 x5
A2 x3
A5 x6
A1 x7

想得到的结果，将A列名称唯一化，出现一个值对应多个值的列表，且想批量处理
A1 x1 x5 x7
A2 x2 x3
A3 x3
A4 x4
A5 x6

解决方案1：perl

use strict;
use warnings;

my %hash;
open OUT, "> lines.txt" or die"$!";

while () {
chomp;
my ($line1,$line2) = split/\s+/;
push @{$hash{$line1}},$line2;
}

foreach my $key(sort keys %hash) {
print OUT "$key\t@{$hash{$key}}\n";
}

close OUT;
__DATA__
A1 x1
A2 x2
A3 x3
A4 x4
A1 x5
A2 x3
A5 x6

A1 x7

解决方案2：R

d = read.table("data.txt")
tapply(d[,2], d[,1], print)