How to replace regex with digit in dataframe

Question

I have a dataframe that includes a series of information that identifies a group of people by their current board memberships. It looks like this:

company_board <- c("company1", "company2", "company3; company 1", "", "")
nonprofit_board <- c("nonprofit1", "", "nonprofit5; nonprofit2", "", "nonprofit3")
df <- data.frame(company_board, nonprofit_board, stringsAsFactors = FALSE)

I want to convert these into simple 1 if there is information in the cell or 0 if there is no data recorded. So for the example I've just given:

company_board <- c("1", "1", "1", "0", "0")
nonprofit_board <- c("1", "0", "1", "0", "1")
df <- data.frame(company_board, nonprofit_board, stringsAsFactors = FALSE)

I know how to use str_extract with [:alnum:] to get the cells that should be 1 but I can't figure out how to then replace these cells with 1 (and the remainder with 0). Any help would be greatly appreciated!

akrun · Accepted Answer · 2020-04-27 18:44:00Z

2

We can also convert to matrix and apply nchar directly

+(nchar(as.matrix(df)) > 0)
#     company_board nonprofit_board
#[1,]             1               1
#[2,]             1               0
#[3,]             1               1
#[4,]             0               0
#[5,]             0               1

answered Apr 27, 2020 at 18:44

akrun

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Comments

ThomasIsCoding · Accepted Answer · 2020-04-27 11:11:30Z

1

An easy way to make it is using nzchar

dfout <- +data.frame(Map(nzchar,df))

such that

> dfout
  company_board nonprofit_board
1             1               1
2             1               0
3             1               1
4             0               0
5             0               1

answered Apr 27, 2020 at 11:11

ThomasIsCoding

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Comments

GKi · Accepted Answer · 2020-04-27 12:38:00Z

1

You can use sapply and test if nchar is more than 0.

sapply(df, function(x) +(nchar(x)>0))
#     company_board nonprofit_board
#[1,]             1               1
#[2,]             1               0
#[3,]             1               1
#[4,]             0               0
#[5,]             0               1

Or shorter:

+(sapply(df, nchar)>0)

In case you have " " you can add something like trimws:

sapply(df, function(x) +(nchar(trimws(x))>0))

Or using nzchar seen already in @ThomasIsCoding's answer.

+sapply(df, nzchar)

edited Apr 27, 2020 at 12:38

answered Apr 27, 2020 at 10:57

GKi

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1 Comment

Oliver Over a year ago

Perfect thanks! I'll accept this answer as soon as the system allows me

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