CODEWITHSUNDEEP
rdataframe
Sciolism Apparently
Sciolism Apparently

Reputation: 355

Creating a table() of counts using a frequency column

I've got data created from the HairEyeColor data

HEC = as.data.frame(HairEyeColor)

which is a quick way of generating a data frame with a Frequency column, which is my situation.

I need to create contingency tables similar to this:

          colhair
coleye black blond brunette red
blue      20    94       84  17
brown     68     7      119  26
green      5    16       29  14
hazel     15    10       54  14

Note: I'm not asking how to do it with the existing HairEyeColor data table, but with a data frame that has a frequency column.

I have tried several varieties of table(), xtabs(), and aggregate() and the best I can do is get is counts of rows. I can't seem to get the frequency column to be productively used.

solutions are not desired.

Upvotes: 2

Views: 876

Answers (2)

M--
M--

Reputation: 29203

We can do it with tapply():

tapply(HEC$Freq, list(ColHair=HEC$Hair,ColEye=HEC$Eye), sum)


  #         ColEye
  # ColHair Brown Blue Hazel Green
  # Black    68   20    15     5
  # Brown   119   84    54    29
  # Red      26   17    14    14
  # Blond     7   94    10    16

Or using data.table package: 

library(data.table)
setDT(HEC)[,list(Freq=sum(Freq)),by=list(Hair, Eye)]

  #     Hair    Eye    Freq
  # 1:  Black   Brown   68
  # 2:  Brown   Brown  119
  # 3:    Red   Brown   26
  # 4:  Blond   Brown    7
  # 5:  Black   Blue    20
  # 6:  Brown   Blue    84
  # 7:    Red   Blue    17
  # 8:  Blond   Blue    94
  # 9:  Black   Hazel   15
  # 10: Brown   Hazel   54
  # 11:   Red   Hazel   14
  # 12: Blond   Hazel   10
  # 13: Black   Green    5
  # 14: Brown   Green   29
  # 15:   Red   Green   14
  # 16: Blond   Green   16

To get it in cross-tab format: 

HEC_tab <- dcast(setDT(HEC)[,list(Freq=sum(Freq)),by=list(Hair, Eye)], 
                                         Hair~Eye, value.var = "Freq")

setnames(HEC_tab , c("HairCol/EyeCol", names(HEC_tab)[-1]))

HEC_tab

  #   HairCol/EyeCol Brown Blue Hazel Green
  # 1:         Black    68   20    15     5
  # 2:         Brown   119   84    54    29
  # 3:           Red    26   17    14    14
  # 4:         Blond     7   94    10    16

Upvotes: 2

akrun
akrun

Reputation: 887781

We do a group by summarise and then spread

library(tidyerse)
HEC %>% 
    group_by(Hair, Eye) %>% 
    summarise(Freq = sum(Freq)) %>%
    spread(Eye, Freq)

It can be also done in a one-liner

xtabs(Freq ~ Eye + Hair, HEC)

Upvotes: 5

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