How to create a new column in data frame with for loop and if statements

Question

I have a data frame with 102 rows, and I need to develop a for loop with an if statement to populate a new column "Season" based on other columns (Sp, Su, Fa, Wi). I have a "1" populating the season that the sample took place (see below).

Sp  Su  Fa  Wi
1   0   0   0
0   0   0   1

I tried just doing summer, in a loop, but I get tons of errors. I just can't seem to grasp For and if loops. ANy help would be appreciated.

for(i in 1:102) {  if(myData$Su==1) myData$Season=Summer}

Error:

In if (myData$Su == 1) myData$Season = Summer :
  the condition has length > 1 and only the first element will be used

Emer · Accepted Answer · 2014-12-10 18:16:09Z

3

Try to identify which column has an 1, then use this index to return the desidered name of the Season from a char vector:

data <- c("Sp  Su  Fa  Wi
           1   0   0   0
           0   0   0   1")
data <- read.table(text=data,header=TRUE)

data$Season <- c("Spring","Summer","Fall","Winter")[which(data==1,arr.ind=TRUE)[,"col"]]

Result:

  Sp Su Fa Wi Season
1  1  0  0  0 Spring
2  0  0  0  1 Winter

answered Dec 10, 2014 at 18:16

Emer

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2 Comments

digEmAll Over a year ago

I like this solution using which ! :)

Emer Over a year ago

You're welcome, note that this is not going to work if there are NA's because which function is going to ignore them and you will have les rows than the data.frame. To avoid this, use the apply version of @Andrie answer

Andrie · Accepted Answer · 2014-12-10 18:23:44Z

1

Since R is a vector-based language, you don't need a for loop in this case.

dat <- data.frame(
  Sp = c(1, 0),
  Su = c(0, 0),
  Fa = c(0, 0),
  Wi = c(0, 1)
)

A naive, brute force way would be to use nested ifelse() functions:

dat$Season <- with(dat, 
                   ifelse(Sp == 1, "Spring", 
                          ifelse(Su == 1, "Summer", 
                                 ifelse(Fa == 1, "Fall", 
                                        "Winter"))))
dat

  Sp Su Fa Wi Season
1  1  0  0  0 Spring
2  0  0  0  1 Winter

But the R way of doing this would be to think about the structure of your data, then use indexing, for example:

dat$season <- apply(dat, 1, function(x) c("Sp", "Su", "Fa", "Wi")[x==1])

  Sp Su Fa Wi season
1  1  0  0  0     Sp
2  0  0  0  1     Wi

edited Dec 10, 2014 at 18:23

answered Dec 10, 2014 at 18:16

Andrie

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1 Comment

MKWalsh Over a year ago

Hi, thank you very much. The first way works fine. The second way I get a column of "NA", but I will continue to fiure out the second way because it seems more concise.

Alexey Ferapontov · Accepted Answer · 2014-12-10 18:14:27Z

0

ifelse(myData$Su==1, myData$Season=="Summer",myData$Season=="Not Summer")

or a more complicated "no" statement (e.g. nested ifelse - if Wi ==1, set to Winter, etc)

answered Dec 10, 2014 at 18:14

Alexey Ferapontov

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Comments

digEmAll · Accepted Answer · 2014-12-10 18:29:48Z

0

If you really want to use a loop you should do in this way :

# recreating an example similar to your data
myData <- read.csv(text= 
"Sp,Su,Fa,Wi
1,0,0,0
0,1,0,0
0,0,1,0
1,0,0,0
0,0,0,1")

# before the loop, add a new "Season" column to myData filled with NAs
myData$Season <- NA

# don't use 102 but nrow(myData) so
# in case myData changes you don't have to modify the code
for(i in 1:nrow(myData)){

  # here you are working row-by-row
  # so note the [i] indexing below

  if(myData$Sp[i] == 1){
    myData$Season[i] = "Spring"
  }else if(myData$Su[i] == 1){
    myData$Season[i] = "Summer"
  }else if(myData$Fa[i] == 1){
    myData$Season[i] = "Fall"
  }else if(myData$Wi[i] == 1){
    myData$Season[i] = "Winter"
  }
}

But actually (as shown in the other answers) there are more efficient and faster ways.

answered Dec 10, 2014 at 18:29

digEmAll

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1 Comment

MKWalsh Over a year ago

THank you very much! all replies have been helpful

akrun · Accepted Answer · 2014-12-10 18:55:59Z

0

You could also use (a variation of @Emer's approach)

 transform(dat, Season=c('Spring', 'Summer', 'Fall',
             'Winter')[as.matrix(seq_len(ncol(dat))*dat)])
 #  Sp Su Fa Wi Season
 #1  1  0  0  0 Spring
 #2  0  0  0  1 Winter

data

 dat <- structure(list(Sp = c(1, 0), Su = c(0, 0), Fa = c(0, 0), Wi = c(0, 
 1)), .Names = c("Sp", "Su", "Fa", "Wi"), row.names = c(NA, -2L
 ), class = "data.frame")

answered Dec 10, 2014 at 18:55

akrun

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