CODEWITHSUNDEEP
rdataframemissing-data
KT_1
KT_1

Reputation: 8474

Delete rows with blank values in one particular column

I am working on a large dataset, with some rows with NAs and others with blanks:

df <- data.frame(ID = c(1:7),                                   
         home_pc = c("","CB4 2DT", "NE5 7TH", "BY5 8IB", "DH4 6PB","MP9 7GH","KN4 5GH"),               
         start_pc = c(NA,"Home", "FC5 7YH","Home", "CB3 5TH", "BV6 5PB",NA),               
         end_pc = c(NA,"CB5 4FG","Home","","Home","",NA))

How do I remove the NAs and blanks in one go (in the start_pc and end_pc columns)? I have in the past used:

df<- df[-which(is.na(df$start_pc)), ]

... to remove the NAs - is there a similar command to remove the blanks?

Upvotes: 79

Views: 236910

Answers (5)

user6074085
user6074085

Reputation: 81

An easy approach would be making all the blank cells NA and only keeping complete cases. You might also look for na.omit examples. It is a widely discussed topic.

df[df==""]<-NA
df<-df[complete.cases(df),]

Upvotes: 8

user6164045
user6164045

Reputation: 81

Alternative solution can be to remove the rows with blanks in one variable: 

df <- subset(df, VAR != "")

Upvotes: 8

Agile Bean
Agile Bean

Reputation: 7141

An elegant solution with dplyr would be:

df %>%
  # recode empty strings "" by NAs
  na_if("") %>%
  # remove NAs
  na.omit

Upvotes: 23

Andrie
Andrie

Reputation: 179398

It is the same construct - simply test for empty strings rather than NA:

Try this:

df <- df[-which(df$start_pc == ""), ]

In fact, looking at your code, you don't need the which, but use the negation instead, so you can simplify it to:

df <- df[!(df$start_pc == ""), ]
df <- df[!is.na(df$start_pc), ]

And, of course, you can combine these two statements as follows:

df <- df[!(df$start_pc == "" | is.na(df$start_pc)), ]

And simplify it even further with with:

df <- with(df, df[!(start_pc == "" | is.na(start_pc)), ])

You can also test for non-zero string length using nzchar.

df <- with(df, df[!(nzchar(start_pc) | is.na(start_pc)), ])

Disclaimer: I didn't test any of this code. Please let me know if there are syntax errors anywhere

Upvotes: 34

sgibb
sgibb

Reputation: 25726

 df[!(is.na(df$start_pc) | df$start_pc==""), ]

Upvotes: 110

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