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I've been working around with this response (that I think is a remarkable well constructed answer) but rather than a single dataframe I have a list of dataframes. So, I've been trying to do this:

head(K[[1]])
        a    b   c
1981 0.76 0.93 NaN
1982  Inf 0.33 NaN
1983 0.25  Inf NaN
1984 0.77 0.73 Inf
1985  Inf 0.85 Inf
1986 0.63 0.56 NaN

K <- lapply(seq_along(K), function(i) for (j in 1:3) set(K[[i]], which(is.infinite(K[[i]][[j]])), j, NA))

Which basically it's supposed to replace the Inf values in every element of the list K but I only get this result:

str(K)

List of 200
 $ : NULL
 $ : NULL
 $ : NULL
 $ : NULL
...

The for loop inside the function works when isolated but I cannot get a list with the infinite-free dataframes.

Any suggestion? Regards.

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noriega
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1 Answers1

3

If K is a list of data frames this produces a new cleaned list of data.frames. The Inf2NA function replaces all infinite values in a vector v with NA.

 Inf2NA <- function(v) replace(v, is.infinite(v), NA)
 lapply(K, function(d) replace(d, TRUE, sapply(d, Inf2NA)))

If it's sufficient to create a list of matrices then this shorter version would be sufficient:

lapply(K, function(d) sapply(d, Inf2NA))
G. Grothendieck
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