如何简单地用NA计算行数-R [英] How to simply count number of rows with NAs - R

问题描述

我试图计算具有整个df的NA的行数，因为我想计算具有df的总行数中具有NA的行的百分比.

I'm trying to compute the number of rows with NA of the whole df as I'm looking to compute the % of rows with NA over the total number of rows of the df.

我已经看过这篇文章:确定具有NA的行数，但它只显示特定范围的列.

I have already have seen this post: Determine the number of rows with NAs but it just shows a specific range of columns.

推荐答案

tl; dr:行明智的，您将需要sum(!complete.cases(DF))或等效的sum(apply(DF, 1, anyNA))

tl;dr: row wise, you'll want sum(!complete.cases(DF)), or, equivalently, sum(apply(DF, 1, anyNA))

有多种方法可以查看数据框中NA值的数量，比例或位置:

There are a number of different ways to look at the number, proportion or position of NA values in a data frame:

其中大多数是从逻辑数据帧开始的，每个NA都带有TRUE，其他所有位置都带有FALSE.对于基本数据集airquality

Most of these start with the logical data frame with TRUE for every NA, and FALSE everywhere else. For the base dataset airquality

is.na(airquality)

此数据集中有44个NA值

sum(is.na(airquality))
# [1] 44

您可以查看每行或每列NA个值的总数:

You can look at the total number of NA values per row or column:

head(rowSums(is.na(airquality)))
# [1] 0 0 0 0 2 1
colSums(is.na(airquality))
#   Ozone Solar.R    Wind    Temp   Month     Day 
 37       7       0       0       0       0 

您也可以使用anyNA()代替is.na():

# by row
head(apply(airquality, 1, anyNA))
# [1] FALSE FALSE FALSE FALSE  TRUE  TRUE
sum(apply(airquality, 1, anyNA))
# [1] 42


# by column
head(apply(airquality, 2, anyNA))
#   Ozone Solar.R    Wind    Temp   Month     Day 
#    TRUE    TRUE   FALSE   FALSE   FALSE   FALSE
sum(apply(airquality, 2, anyNA))
# [1] 2

complete.cases()可以使用，但只能逐行使用:

complete.cases() can be used, but only row-wise:

sum(!complete.cases(airquality))
# [1] 42

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