如何在R的仅两列中省略带有NA的行? [英] How to omit rows with NA in only two columns in R?

问题描述

我想省略两列的两者中的NA出现的行.

I want to omit rows where NA appears in both of two columns.

我熟悉na.omit，is.na和complete.cases，但是无法弄清楚如何使用它们来获得我想要的东西.例如，我有以下数据框:

I'm familiar with na.omit, is.na, and complete.cases, but can't figure out how to use these to get what I want. For example, I have the following dataframe:

(df <- structure(list(x = c(1L, 2L, NA, 3L, NA),
                     y = c(4L, 5L, NA, 6L, 7L),
                     z = c(8L, 9L, 10L, 11L, NA)),
                .Names = c("x", "y", "z"),
                class = "data.frame",
                row.names = c(NA, -5L)))
x   y   z
1   4   8
2   5   9
NA  NA  10
3   6   11
NA  7   NA

并且我只想删除 那些在x和y列中均出现NA的行(不包括z中的任何内容)，以给出

and I want to remove only those rows where NAappears in both the x and y columns (excluding anything in z), to give

x   y   z
1   4   8
2   5   9
3   6   11
NA  7   NA

有人知道这样做的简单方法吗?使用na.omit，is.na或complete.cases不起作用.

Does anyone know an easy way to do this? Using na.omit, is.na, or complete.cases is not working.

推荐答案

df[!with(df,is.na(x)& is.na(y)),]
#      x y  z
#1  1 4  8
#2  2 5  9
#4  3 6 11
#5 NA 7 NA

我确实在稍大的数据集上进行了基准测试.结果如下:

I did benchmarked on a slightly bigger dataset. Here are the results:

set.seed(237)
df <- data.frame(x=sample(c(NA,1:20), 1e6, replace=T), y= sample(c(NA, 1:10), 1e6, replace=T), z= sample(c(NA, 5:15), 1e6,replace=T)) 

f1 <- function() df[!with(df,is.na(x)& is.na(y)),]
f2 <- function() df[rowSums(is.na(df[c("x", "y")])) != 2, ]
f3 <- function()  df[ apply( df, 1, function(x) sum(is.na(x))>1 ), ] 

library(microbenchmark)

microbenchmark(f1(), f2(), f3(), unit="relative")
Unit: relative
#expr       min        lq    median        uq       max neval
# f1()  1.000000  1.000000  1.000000  1.000000  1.000000   100
# f2()  1.044812  1.068189  1.138323  1.129611  0.856396   100
# f3() 26.205272 25.848441 24.357665 21.799930 22.881378   100

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