Question

我有以下代码...

pMiss <- function(x){sum(is.na(x))}
missing_values_data <- apply(csv,1,pMiss)

但这只会输出类似...

[1]    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0
[2]    0    0    0    0    0    0    0    0   20    0    0    0    0    0    0    0    0
[3]    0    0    0    0   10    0    0    0    0    0    0    0    0    0    0    0    0
[4]    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0
[5]    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0    0

10和20是包含NA的行。有没有一种方法可以输出这些行在数据帧中的索引位置，例如[3,5]等。

Answer 1

sapply(df, function(x){which(is.na(x))})

# Data Used: 

df <- data.frame(x1 = c(84.2,105.2,79.2,NA,108.2,79.2,112.2,NA,114.2,92.2),
             x2 = c(138.3,110.3,84.3,45.3,128.3,99.3,NA,124.3,121.3,115.3),
             x3 = as.character(c(138.3,110.3,84.3,45.3,128.3,99.3,100.3,124.3,121.3,115.3)),
             stringsAsFactors = FALSE)

Answer 2

您可以在which函数中使用参数arr.ind返回数组索引。简短示例

df <- as.data.frame(matrix(rnorm(1000), ncol = 10, nrow = 100))
df[1,3] <- NA
df[3,9] <- NA

which(is.na(df), arr.ind = TRUE)

     row col
[1,]   1   3
[2,]   3   9

我想念这个问题，如果您只对行感兴趣，那么只需提取正确的列即可

  > which(is.na(df), arr.ind = TRUE)[ ,1]
[1] 1 3

如何在数据帧中输出NA值的位置？

2 个答案: