using a for-loop to reassign selected values within a data frame

[Bonnie Dixon]

This seems like it should be simple, so I'm not sure why I am having trouble.  

I have a data frame of character strings, like this one, in which all values should say "ab", but some of the b's are missing, so I need to correct those values.  

df1 <- data.frame(A=c("ab", "a", "ab"), B=c("a", "ab", "a"))

Luckily, I have a list of the row indices where these errors occur in each column of the data frame.  

l1 <- list(A=2, B=c(1,3))

I tried to write a for-loop to iterate through the columns and paste in the necessary "b" at each of the rows indicated by the list:

for(i in c(1:2)) {

  df1$i[l1$i] <- paste0(df1$i[l1$i], "b")

}

But my for-loop is returning an error:

Error in `$<-.data.frame`(`*tmp*`, "i", value = character(0)) : 

  replacement has 0 rows, data has 3

What am I doing wrong?

Bonnie

[Bonnie Dixon]

Opps.  That for-loop should be:

for(i in names(l1)) {

  df1$i[l1$i] <- paste0(df1$i[l1$i], "b")

}

Bonnie

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[Vince S. Buffalo]

Hi Bonnie,

I might just use ifelse here:

> d <- data.frame(A=c("a", "ab", "a", "a", "c"), stringsAsFactors=FALSE)

> d$Afixed <- ifelse(d$A == "a", "ab", d$A)

> d

   A Afixed

1  a     ab

2 ab     ab

3  a     ab

4  a     ab

5  c      c

Hope this helps,

Vince

--

Vince Buffalo

Ross-Ibarra Lab (www.rilab.org)

Plant Sciences, UC Davis

[Bonnie Dixon]

That approach still returned an error when I tried to automate it within a for-loop.  (The reason why I am trying to automate this process is because the real data frame that I need to do this manipulation on has many columns that need to be changed and also some columns that I don't want to change.)

 df1 <- data.frame(A=c("ab", "a", "ab"), B=c("a", "ab", "a"),

                  stringsAsFactors=F)

 l1 <- list(A=2, B=c(1,3))

 for(i in names(l1)) {

 df1$i <- ifelse(df1$i=="a", "ab", df1$i)

 }

 

 Error in `$<-.data.frame`(`*tmp*`, "i", value = logical(0)) : 

  replacement has 0 rows, data has 3 

But, I have figured out what was wrong.  Apparently using $ indexing was the problem, because when I switch to bracket indexing within the for-loop, either this approach, or the original approach that I tried, works fine.

for(i in names(l1)) {

  df1[ , i] <- ifelse(df1[ , i]=="a", "ab", df1[ , i])

}

for(i in names(l1)) {

  df1[l1[[i]], i] <- paste0(df1[l1[[i]], i], "b")

}

The same seems to be true for the apply() statements I had tried.  They work with bracket indexing, but not with the use of the dollar sign operator.  If anyone can explain why the $ doesn't work within a for-loop or apply() statement, I'd love to know the reason.

Bonnie

[Vince S. Buffalo]

Hi Bonnie,

Sorry, I missed that you needed to apply this to many columns. Yes, dollar signs accessor will not work for this because it's interpreting the i literally — looking for a column named "i". You do need to use square brackets if i is a variable pointing to a column name as a string/character vector.

Vince

[Vince S. Buffalo]

Also, I think this might be the cleanest way to do it:

>  d <- data.frame(A=c("a", "ab", "a", "a", "c"), B=c("a", "x", "ab", "a", "a"), C=runif(5), stringsAsFactors=FALSE)

> bad_cols <- c("A", "B")

> dcopy <- d

> dcopy[bad_cols] <- lapply(dcopy[bad_cols], function(x) ifelse(x == "a", "ab", x))

> dcopy

   A  B          C

1 ab ab 0.44395754

2 ab  x 0.04420908

3 ab ab 0.25378545

4 ab ab 0.99977372

5  c ab 0.06727521

I've used a copy of the original dataframe so it could be compared to the original (since this overwrites existing columns).

This solution exploits the fact that data.frames are just lists under the hood. They need to be lists, because data.frames can have columns of heterogeneous data type and lists are R's primary structure for storing data with heterogeneous type. 

> is.list(d)

[1] TRUE

HTH,

Vince

[Bonnie Dixon]

Yes, that is a great way to do it.  For my real task I needed to replace the ifesle() statement with gsub() and use a regular expression, but I got it to work, and that's what counts!

Thanks also for the explanation regarding the $ accessor, Vince.  That will be easier to remember now that I know the reason why it didn't work.

Bonnie