CODEWITHSUNDEEP
r
Bfu38
Bfu38

Reputation: 1141

Replace values in columns of a data frame after match a pattern in another column

Is there an intuitive way to replace values in two columns of a data frame if patterns in one column of the same differ from the desired one? Specifically, my data frame looks like:

   Col1              Value1      Value2        
   my_HD_SAMPLE      34          0.34        
   my_HD_T_SAMPLE    6           0.6           
   my_BD_NAME        94          0.94      
   my_LE_NAME        1           0.1  
   my_TS_SAMPLE      74          0.74

I would like to replace the values in columns 2 and 3 with 0 if the row in Col1 does not match the pattern BD|HD. The desired output should be:

   Col1              Value1     Value2        
   my_HD_SAMPLE      34         0.34        
   my_HD_T_SAMPLE    6          0.6           
   my_BD_NAME        94         0.94      
   my_LE_NAME        0          0  
   my_TS_SAMPLE      0          0

I tried this:

mydf = mydf$"Col1"[!grepl('BD|HD',mydf$"Col1")] <- 0

And I got this warning message:

In [<-.factor`(`*tmp*`, !grepl("BD|HD", mydf$Col1), value = c(`NA` = NA,  :
  invalid factor level, NA generated

How can I solve this problem?

Upvotes: 0

Views: 784

Answers (2)

Basil Eric Rabi
Basil Eric Rabi

Reputation: 343

Seems to me Col1 is a factor. Try:

# Convert to character first.
rows <- !grepl("BD|HD", as.character(mydf$`Col1`))
mfdf$`Value1`[rows] <- 0
mfdf$`Value2`[rows] <- 0

Upvotes: 1

Mohanasundaram
Mohanasundaram

Reputation: 2949

Try This

mydf[!grepl('BD|HD',mydf$"Col1"),][2:ncol(mydf)] <- 0

> mydf
            Col1 Value1 Value2
1   my_HD_SAMPLE     34   0.34
2 my_HD_T_SAMPLE      6   0.60
3     my_BD_NAME     94   0.94
4     my_LE_NAME      0   0.00
5   my_TS_SAMPLE      0   0.00

Upvotes: 0

Related Questions