CODEWITHSUNDEEP
rtidyr
dotsbyname
dotsbyname

Reputation: 250

reshaping factor data to matrix

What is a correct/short way to reshape a factor column in data:

login has_profile_in
1234  Facebook
1234  LinkedIn
1235  VK

into a matrix like this:

login Facebook LinkedIn VK
1234     1        1      0
1235     0        0      1

using tidyr pipeline?

Edit: I know some regular ways of doing this, i.e. with reshape2

dcast(login~has_profile_in)

and that there are other ways as well. My question is how to do it in a tidyr way, including the operation in a general pipeline-based framework

Upvotes: 0

Views: 112

Answers (3)

akrun
akrun

Reputation: 887501

As the OP requested tidyr method

library(dplyr)
library(tidyr)
df1 %>% 
   mutate(Count = 1) %>%
   spread(has_profile_in, Count, fill = 0)
#   login Facebook LinkedIn VK
#1  1234        1        1  0
#2  1235        0        0  1

The shortest option would be table

 as.data.frame.matrix(+(table(df1)!=0))
 #     Facebook LinkedIn VK
 #1234        1        1  0
 #1235        0        0  1

Or using data.table

library(data.table)
dcast(setDT(df1), login~has_profile_in, function(x) +(length(x)!=0))
#    login Facebook LinkedIn VK
#1:  1234        1        1  0
#2:  1235        0        0  1

NOTE: dcast would be the fastest but biased voting is still going on or sockpuppet activity.

Upvotes: 1

Ronak Shah
Ronak Shah

Reputation: 389135

You can use aggregate

aggregate(has_profile_in ~ login, df, table)


#   login has_profile_in.Facebook has_profile_in.LinkedIn has_profile_in.VK
#1  1234                       1                       1                 0
#2  1235                       0                       0                 1

You can rename the columns using setNames and make it more readable

setNames(aggregate(has_profile_in ~ login, df, table), c("Login", ""))

# Login   .Facebook .LinkedIn .VK
#1  1234         1         1   0
#2  1235         0         0   1

Upvotes: 3

jkt
jkt

Reputation: 946

You can use

model.matrix(~yourFactor+0)

Upvotes: 1

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