R: Convert factor column to multiple boolean columns

Question

I am trying to convert a factor column into multiple boolean columns as the image below shows. The data is from weather stations as retrieved using the fine weatherData package. The factor column I want to convert into multiple boolean columns contains 11 factors. Some of them are single "events", and some of them are a combination of "events".

Here is an image showing what I want to achieve: enter image description here This is R code which will produce the data frame with combined factors that I want to convert into several boolean columns:

df <- read.table(text="
date    Events
1/8/2013    Rain
1/9/2013    Fog
1/10/2013   ''
1/11/2013   Fog-Rain
1/12/2013   Snow
1/13/2013   Rain-Snow
1/14/2013   Rain-Thunderstorm
1/15/2013   Thunderstorm
1/16/2013   Fog-Rain-Thunderstorm
1/17/2013   Fog-Thunderstorm
1/18/2013   Fog-Rain-Thunderstorm-Snow",
                 header=T)
df$date <- as.character(as.Date(df$date, "%m/%d/%Y"))

Thanks in advance.

akrun · Accepted Answer · 2014-08-17 14:23:22Z

13

You could try:

 lst <- strsplit(as.character(df$Events),"-")
 lvl <- unique(unlist(lst))      
 res <- data.frame(date=df$date,
            do.call(rbind,lapply(lst, function(x) table(factor(x, levels=lvl)))), 
                                       stringsAsFactors=FALSE)

  res
 #         date Rain Fog Snow Thunderstorm
 #1  2013-01-08    1   0    0            0
 #2  2013-01-09    0   1    0            0
 #3  2013-01-10    0   0    0            0
 #4  2013-01-11    1   1    0            0
 #5  2013-01-12    0   0    1            0
 #6  2013-01-13    1   0    1            0
 #7  2013-01-14    1   0    0            1
 #8  2013-01-15    0   0    0            1
 #9  2013-01-16    1   1    0            1
 #10 2013-01-17    0   1    0            1
# 11 2013-01-18    1   1    1            1

Or possibly, this could be faster than the above (contributed by @alexis_laz)

  setNames(data.frame(df$date, do.call(rbind,lapply(lst, function(x) as.integer(lvl %in% x)) )), c("date", lvl))

Or

 library(devtools)
 library(data.table)
 source_gist("11380733")
 library(reshape2) #In case it is needed 

 res1 <- dcast.data.table(cSplit(df, "Events", "-", "long"), date~Events)
 res2 <- merge(subset(df, select=1), res1, by="date", all=TRUE)
 res2 <- as.data.frame(res2)
 res2[,-1]  <- (!is.na(res2[,-1]))+0
 res2[,c(1,3,2,4,5)]
 #          date Rain Fog Snow Thunderstorm
  #1  2013-01-08    1   0    0            0
  #2  2013-01-09    0   1    0            0
  #3  2013-01-10    0   0    0            0
  #4  2013-01-11    1   1    0            0
  #5  2013-01-12    0   0    1            0
  #6  2013-01-13    1   0    1            0
  #7  2013-01-14    1   0    0            1
  #8  2013-01-15    0   0    0            1
  #9  2013-01-16    1   1    0            1
  #10 2013-01-17    0   1    0            1
  #11 2013-01-18    1   1    1            1

Or

 library(qdap)
 with(df, termco(Events, date, c("Rain", "Fog", "Snow", "Thunderstorm")))[[1]][,-2]
 #         date Rain Fog Snow Thunderstorm
 #1  2013-01-08    1   0    0            0
 #2  2013-01-09    0   1    0            0
 #3  2013-01-10    0   0    0            0
 #4  2013-01-11    1   1    0            0
 #5  2013-01-12    0   0    1            0
 #6  2013-01-13    1   0    1            0
 #7  2013-01-14    1   0    0            1
 #8  2013-01-15    0   0    0            1
 #9  2013-01-16    1   1    0            1
 #10 2013-01-17    0   1    0            1
 #11 2013-01-18    1   1    1            1

edited Aug 17, 2014 at 14:23

answered Aug 17, 2014 at 9:15

akrun

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18 Comments

Spacedman Over a year ago

Second example needs reshape2 for dcast

akrun Over a year ago

@Spacedman. Thanks. I thought dcast alone from data.table would work as I loaded both reshape2 and data.table. Seems like dcast.data.table is the one that works without using reshape2

akrun Over a year ago

@David Arenburg. I tried it without loading reshape2 and it is working for me.

akrun Over a year ago

@David Arenburg. Mine is data.table_1.9.2

akrun Over a year ago

@David Arenburg. I tried it again on a fresh console. After library(data.table), checked sessionInfo() loaded via a namespace (and not attached): [1] plyr_1.8.1 Rcpp_0.11.1 reshape2_1.4 stringr_0.6.2

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A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2014-08-17 16:24:26Z

The easiest thing I can think of is concat.split.expanded from my "splitstackshape" package (devel version 1.3.0, from GitHub).

## Get the right version of the package
library(devtools)
install_github("splitstackshape", "mrdwab", ref = "devel")
packageVersion("splitstackshape")
# [1] ‘1.3.0’

## Split up the relevant column
concat.split.expanded(df, "Events", "-", type = "character", 
                      fill = 0, drop = TRUE)
#          date Events_Fog Events_Rain Events_Snow Events_Thunderstorm
# 1  2013-01-08          0           1           0                   0
# 2  2013-01-09          1           0           0                   0
# 3  2013-01-10          0           0           0                   0
# 4  2013-01-11          1           1           0                   0
# 5  2013-01-12          0           0           1                   0
# 6  2013-01-13          0           1           1                   0
# 7  2013-01-14          0           1           0                   1
# 8  2013-01-15          0           0           0                   1
# 9  2013-01-16          1           1           0                   1
# 10 2013-01-17          1           0           0                   1
# 11 2013-01-18          1           1           1                   1

Answering this question, I realize that I've somewhat foolishly hard-coded a "trim" feature in concat.split.expanded that could slow things down a lot. If you want a much faster approach, use charMat (the function called by concat.split.expanded) directly on the split up version of your "Events" column, like this:

splitstackshape:::charMat(
    strsplit(as.character(indf[, "Events"]), "-", fixed = TRUE), fill = 0)

For some benchmarks, check out this Gist.

Tyler Rinker · Accepted Answer · 2014-08-17 12:27:14Z

Here's an approach with qdapTools:

library(qdapTools)

matrix2df(mtabulate(lapply(split(as.character(df$Events), df$date), 
    function(x) strsplit(x, "-")[[1]])), "Date")

##          Date Fog Rain Snow Thunderstorm
## 1  2013-01-08   0    1    0            0
## 2  2013-01-09   1    0    0            0
## 3  2013-01-10   0    0    0            0
## 4  2013-01-11   1    1    0            0
## 5  2013-01-12   0    0    1            0
## 6  2013-01-13   0    1    1            0
## 7  2013-01-14   0    1    0            1
## 8  2013-01-15   0    0    0            1
## 9  2013-01-16   1    1    0            1
## 10 2013-01-17   1    0    0            1
## 11 2013-01-18   1    1    1            1

Here is the same answer with magrittr as it makes the chain clearer:

split(as.character(df$Events), df$date) %>%
    lapply(function(x) strsplit(x, "-")[[1]]) %>%
    mtabulate() %>%
    matrix2df("Date")

rnso · Accepted Answer · 2014-08-17 12:48:17Z

4

Can be done with base R using 'grep':

ddf = data.frame(df$date, df$Events, "Rain"=rep(0), "Fog"=rep(0), "Snow"=rep(0), "Thunderstorm"=rep(0)) 

for(i in 3:6)   ddf[grep(names(ddf)[i],ddf[,2]),i]=1

ddf
      df.date                  df.Events Rain Fog Snow Thunderstorm
1  2013-01-08                       Rain    1   0    0            0
2  2013-01-09                        Fog    0   1    0            0
3  2013-01-10                               0   0    0            0
4  2013-01-11                   Fog-Rain    1   1    0            0
5  2013-01-12                       Snow    0   0    1            0
6  2013-01-13                  Rain-Snow    1   0    1            0
7  2013-01-14          Rain-Thunderstorm    1   0    0            1
8  2013-01-15               Thunderstorm    0   0    0            1
9  2013-01-16      Fog-Rain-Thunderstorm    1   1    0            1
10 2013-01-17           Fog-Thunderstorm    0   1    0            1
11 2013-01-18 Fog-Rain-Thunderstorm-Snow    1   1    1            1

edited Aug 17, 2014 at 12:48

answered Aug 17, 2014 at 12:41

rnso

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2 Comments

A5C1D2H2I1M1N2O1R2T1 Over a year ago

+1. I'm just updating my benchmarks and this is a speedy solution (the fastest so far) if you know in advance the values to expect.

Rumpleteaser Over a year ago

I've built on this answer to dynamically identify the unique options so that you don't need to know the values in advance.

Community · Accepted Answer · 2017-05-23 12:00:44Z

2

Create a vector with factors

set.seed(1)
n <- c("Rain", "Fog", "Snow", "Thunderstorm")
v <- sapply(sample(0:3,100,T), function(i) paste0(sample(n,i), collapse = "-"))
v <- as.factor(v)

Function which returns matrix with desired output that shoulb be cbind'ed to the initial data.frame

mSplit <- function(vec) {
  if (!is.character(vec))
    vec <- as.character(vec)
  L <- strsplit(vec, "-")
  ids <- unlist(lapply(seq_along(L), function(i) rep(i, length(L[[i]])) ))
  U <- sort(unique(unlist(L)))
  M <- matrix(0, nrow = length(vec), 
              ncol = length(U), 
              dimnames = list(NULL, U))
  M[cbind(ids, match(unlist(L), U))] <- 1L
  M
}

Solution is based on the answer of Ananda Mahto to that SO question. It should be pretty fast.

res <- mSplit(v)

edited May 23, 2017 at 12:00

CommunityBot

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answered Aug 17, 2014 at 16:51

DrDom

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1 Comment

A5C1D2H2I1M1N2O1R2T1 Over a year ago

+1. This is essentially what you would see in the code for charMat from "splitstackshape" mentioned in my answer.

Rshad Zhran · Accepted Answer · 2020-09-10 13:10:39Z

2

I think what you need in this case is a simple call for the function dummy. Let's call the target column. target_cat.

df_target_bin <- data.frame(dummy(target_cat, "<prefix>"))

This will create a new data frame with a column with 0s and 1s values for each value of target_cat.

To convert the columns into logical, and with logical I mean the values be TRUE and FALSE, then use the function as.logical.

df_target_logical <- apply(df_target_bin, as.logical)

answered Sep 10, 2020 at 13:10

Rshad Zhran

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Comments

Rumpleteaser · Accepted Answer · 2020-10-18 11:00:32Z

2

Building on the answer by @rnso

The following will identify all the unique elements and then dynamically generate new columns with the relevant data in them.

options = unique(unlist(strsplit(df$Events, '-'), recursive=FALSE))
for(o in options){
  df$newcol = rep(0)
  df <- rename(df, !!o := newcol)
  df[grep(o, df$Events), o] = 1
}

Results:

         date                     Events Rain Fog Snow Thunderstorm
1  2013-01-08                       Rain    1   0    0            0
2  2013-01-09                        Fog    0   1    0            0
3  2013-01-10                               0   0    0            0
4  2013-01-11                   Fog-Rain    1   1    0            0
5  2013-01-12                       Snow    0   0    1            0
6  2013-01-13                  Rain-Snow    1   0    1            0
7  2013-01-14          Rain-Thunderstorm    1   0    0            1
8  2013-01-15               Thunderstorm    0   0    0            1
9  2013-01-16      Fog-Rain-Thunderstorm    1   1    0            1
10 2013-01-17           Fog-Thunderstorm    0   1    0            1
11 2013-01-18 Fog-Rain-Thunderstorm-Snow    1   1    1            1

answered Oct 18, 2020 at 11:00

Rumpleteaser

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1 Comment

rnso Over a year ago

Good modification!

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