Replace strings in data frame columns with integer in R

Question

I have a data frame called 'foo':

 foo <- data.frame("row1" = c(1,2,3,4,5), "row2" = c(1,2.01,3,"-","-"))

'foo' was uploaded from a different program as a CSV file and has two columns. one is a numerical data type and the other is a factor data type.

str(foo)
'data.frame':   5 obs. of  2 variables:
$ row1: num  1 2 3 4 5
$ row2: Factor w/ 4 levels "-","1","2.01",..: 2 3 4 1 1

Notice there are dashes, e.g. "-" , in foo$row2, which causes this column to be a factor. I want to replace the dashes with zeros, such that data.class(foo$row2) will return 'numerical'. The idea is to replace all dashes in each column so I can run numberical analyses on it with R.

What is the simplest way to do this in R?

Thanks,

Metrics · Accepted Answer · 2013-10-17 22:04:29Z

2

Q: The idea is to replace all dashes in each column so I can run numerical analyses on it with R.

Use apply or sapply with sub

 kk<-data.frame(apply(foo,2,function(x) as.numeric(sub("-",0,x))))
> kk
  row1 row2
1    1 1.00
2    2 2.01
3    3 3.00
4    4 0.00
5    5 0.00

> str(kk$row2)
 num [1:5] 1 2.01 3 0 0

Or, you can use sapply

kk<-data.frame(sapply(names(foo),function(x)as.numeric(sub("-",0,foo[,x]))))

Update: If you want just the second col, you don't need to use apply:foo$row2<- as.numeric(sub("-",0,foo[,2]))

edited Oct 17, 2013 at 22:04

answered Oct 17, 2013 at 21:23

Metrics

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4 Comments

AME Over a year ago

Does this apply the string replace function to each column in the data frame? If so, how can I target the second column only? Thx!

Metrics Over a year ago

Yes, it does for all columns. If you want just the second col, you don't need to use apply:foo$row2<- as.numeric(sub("-",0,foo[,2]))

AME Over a year ago

Instead of calling foo[,2] for the second column index, how can I call it by column name, i.e. foo$row2<- as.numeric(sub("-",0,foo[,foo$row2]))?

Simon O'Hanlon Over a year ago

@AME did you even look at the other answers? This is exactly what I posted.

mrip · Accepted Answer · 2013-10-17 21:19:15Z

2

Here is one simple way to do it. There might be a more elegant way, but this will work:

> foo <- data.frame("row1" = c(1,2,3,4,5), "row2" = c(1,2.01,3,"-","-"))
> levels(foo$row2)[levels(foo$row2)=="-"]<-0
> foo$row2<-as.numeric(as.character(foo$row2))
> class(foo$row2)
[1] "numeric"
> foo
  row1 row2
1    1 1.00
2    2 2.01
3    3 3.00
4    4 0.00
5    5 0.00

answered Oct 17, 2013 at 21:19

mrip

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Comments

Stedy · Accepted Answer · 2013-10-17 21:17:48Z

1

I would use ifelse() for this: foo$row2 <- ifelse(foo$row2 == "-", 0, as.numeric(foo$row2))

you might also need to as as.character() to convert from factor to character

answered Oct 17, 2013 at 21:17

Stedy

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2 Comments

TheComeOnMan Over a year ago

And an as.numeric to convert it to the numeric form OP needs.

AME Over a year ago

Running this code on the real data set (not the example one shown here) returns #NA Coerced. The real data set that i'm trying to run this function on contains commas, e.g. 1,000. This seems to force an #NA coercion with the command you provided.

Simon O'Hanlon · Accepted Answer · 2013-10-17 21:34:07Z

1

How about gsub...

as.numeric( gsub("-" , 0 , foo[,2] ) )
#[1] 1.00 2.01 3.00 0.00 0.00

answered Oct 17, 2013 at 21:34

Simon O'Hanlon

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