Contingency table once against multiple table

Question

I have a table like the following:

V1  V2  V3  V4  V5  V6  V7  V8  V9 V10
fve fve fve fve fve fve fve fve mdm mdm
mdm fve fve fve fve fve fve fve fve fve
fve fve fve fve fve fve fve fve fve fve
mdm fve fve fve fve fve fve fve fve fve
fve fve fve fve fve fve fve fve fve fve

and I would like to count the frequence from column 1 against all and get something like this:

    fve mdm
fve 25  2
mdm 18  0

Is it possible to do it ? I tried with table() and ftable() but output is not corresponding to what I'm expecting.

Please explain how you got the numbers in your results from your original table. — BenBarnes, Jun 03 '15 at 06:08
example line 1: fve fve fve fve fve fve fve fve mdm mdm you have 2 variable fve and mdm. The references is column one so you have : fve vs fve = 7 and fve vs mdm = 2 — Aurore Vaitinadapoule, Jun 03 '15 at 06:10

IRTFM · Answer 1 · 2015-06-03T06:20:01.443

3

For the modified question:

> tapply(unlist(dat[-1]), rep(dat[[1]], length(dat[-1])), table)
$fve

fve mdm 
 25   2 

$mdm

fve mdm 
 18   0

I suppose you could run do.call(rbind, ...) on those values to get the desired table:

edited Jun 03 '15 at 06:20

answered Jun 03 '15 at 05:44

IRTFM

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This will wrapp the columns. It's not exactly what I mean. Please the edition of my question. – Aurore Vaitinadapoule Jun 03 '15 at 05:58
2

You should always provide a data object that has the needed complexity for testing. – IRTFM Jun 03 '15 at 06:19

score 2 · Accepted Answer · answered Jun 03 '15 at 06:38

2

Slightly different option is

table(df1[,1][row(df1[-1])], unlist(df1[-1])) 
#     fve mdm
# fve  25   2
# mdm  18   0

answered Jun 03 '15 at 06:38

akrun

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Contingency table once against multiple table

2 Answers2