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从R数据帧列表中清除Inf值

[英]Cleaning Inf values from an R list of dataframes

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. 我一直在处理此响应 (我认为这是一个了不起的构造合理的答案),但是我没有一个数据框,而是有一个数据框列表。 So, I've been trying to do this: 因此,我一直在尝试这样做:

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: 基本上应该替换列表K的每个元素中的Inf值,但我只能得到以下结果:

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. 孤立时该函数内部的for循环有效,但我无法获得包含无穷大数据帧的列表。

Any suggestion? 有什么建议吗? Regards. 问候。

If K is a list of data frames this produces a new cleaned list of data.frames. 如果K是数据帧列表,则将生成一个新的data.frames清除列表。 The Inf2NA function replaces all infinite values in a vector v with NA. Inf2NA函数用NA替换向量v中的所有无限值。

 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: 如果创建一个矩阵列表就足够了,那么这个简短的版本就足够了:

lapply(K, function(d) sapply(d, Inf2NA))

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