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删除矩阵中的NA / NaN / Inf

[英]Remove NA/NaN/Inf in a matrix

I want to try two things : 我想尝试两件事:

  1. How do I remove rows that contain NA/NaN/Inf 如何删除包含NA / NaN / Inf的行
  2. How do I set value of data point from NA/NaN/Inf to 0. 如何将数据点的值从NA / NaN / Inf设置为0。

So far, I have tried using the following for NA values, but been getting warnings. 到目前为止,我已尝试将以下内容用于NA值,但已收到警告。

> eg <- data[rowSums(is.na(data)) == 0,]
  Error in rowSums(is.na(data)) : 'x' must be an array of at least two dimensions In addition: Warning message: In is.na(data) : is.na() applied to non-(list or vector) of type 'closure' 

I guess I'll throw my hat into the ring with my preferred methods: 我想我会用我喜欢的方法把帽子扔进戒指:

# sample data
m <- matrix(c(1,2,NA,NaN,1,Inf,-1,1,9,3),5)
# remove all rows with non-finite values
m[!rowSums(!is.finite(m)),]
# replace all non-finite values with 0
m[!is.finite(m)] <- 0
library(functional)
m[apply(m, 1, Compose(is.finite, all)),]

Demonstration: 示范:

m <- matrix(c(1,2,3,NA,4,5), 3)
m
##      [,1] [,2]
## [1,]    1   NA
## [2,]    2    4
## [3,]    3    5

m[apply(m, 1, Compose(is.finite, all)),]
##      [,1] [,2]
## [1,]    2    4
## [2,]    3    5

Note: Compose(is.finite, all) is equivalent to function(x) all(is.finite(x)) 注意: Compose(is.finite, all)等效于function(x) all(is.finite(x))

To set the values to 0, use matrix indexing: 要将值设置为0,请使用矩阵索引:

m[!is.finite(m)] <- 0
m
##      [,1] [,2]
## [1,]    1    0
## [2,]    2    4
## [3,]    3    5

NaRV.omit(x) is my preferred option for question 1. Mnemonic NaRV means "not a regular value". NaRV.omit(x)是问题1的首选选项。助记符NaRV表示“不是常规值”。

require(IDPmisc)
m <- matrix(c(1,2,3,NA,5, NaN, 7, 8, 9, Inf, 11, 12, -Inf, 14, 15), 5)
> m
     [,1] [,2] [,3]
[1,]    1  NaN   11
[2,]    2    7   12
[3,]    3    8 -Inf
[4,]   NA    9   14
[5,]    5  Inf   15
> NaRV.omit(m)
     [,1] [,2] [,3]
[1,]    2    7   12
attr(,"na.action")
[1] 1 3 4 5
attr(,"class")
[1] "omit"

Just another way (for the first question): 只是另一种方式(对于第一个问题):

m <- structure(c(1, 2, 3, NA, 4, 5, Inf, 5, 6, NaN, 7, 8), 
              .Dim = c(4L, 3L))
#      [,1] [,2] [,3]
# [1,]    1    4    6
# [2,]    2    5  NaN
# [3,]    3  Inf    7
# [4,]   NA    5    8

m[complete.cases(m * 0), , drop=FALSE]
#      [,1] [,2] [,3]
# [1,]    1    4    6

I can't think anything else other than Matthew's answer for the second part. 除了马修对第二部分的回答,我想不出别的什么。

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