Say I write the following code to produce a dataframe:
假设我编写以下代码来生成数据帧:
name <- c("Joe","John","Susie","Mack","Mo","Curly","Jim")
age <- c(1,2,3,NaN,4,5,NaN)
DOB <- c(10000, 12000, 16000, NaN, 18000, 20000, 22000)
DOB <- as.Date(DOB, origin = "1960-01-01")
trt <- c(0, 1, 1, 2, 2, 1, 1)
df <- data.frame(name, age, DOB, trt)
that looks like this:
看起来像这样:
name age DOB trt
1 Joe 1 1987-05-19 0
2 John 2 1992-11-08 1
3 Susie 3 2003-10-22 1
4 Mack NaN <NA> 2
5 Mo 4 2009-04-13 2
6 Curly 5 2014-10-04 1
7 Jim NaN 2020-03-26 1
How would I be able to remove rows where both age and DOB have missing values for the row? For example, I'd like a new dataframe (df2) to look like this:
如何删除年龄和DOB都缺少该行的值的行?例如,我想要一个新的数据帧(df2)看起来像这样:
name age DOB trt
1 Joe 1 1987-05-19 0
2 John 2 1992-11-08 1
3 Susie 3 2003-10-22 1
5 Mo 4 2009-04-13 2
6 Curly 5 2014-10-04 1
7 Jim NaN 2020-03-26 1
I've tried the following code, but it deleted too many rows:
我尝试了以下代码,但它删除了太多行:
df2 <- df[!(is.na(df$age)) & !(is.na(df$DOB)), ]
In SAS, I would just write WHERE missing(age) ge 1 AND missing(DOB) ge 1
in a DATA step, but obviously R has different syntax.
在SAS中,我只会在DATA步骤中写入WHERE missing(age)ge 1 AND missing(DOB)ge 1,但显然R具有不同的语法。
Thanks in advance!
提前致谢!
2 个解决方案
#1
1
If you want to remove those rows where two columns (age and DOB) have more than 1 NA (which would mathematically mean that there could only be 2 NAs in such a case), you can do for example:
如果要删除两列(年龄和DOB)具有多于1个NA的行(这在数学上意味着在这种情况下只能有2个NA),您可以执行以下操作:
df[!is.na(df$age) | !is.na(df$DOB),]
which means that either both or one of the columns should be not NA, or
这意味着两个列或其中一个列不应该是NA,或者
df[rowSums(is.na(df[2:3])) < 2L,]
which means that the sum of NAs in columns 2 and 3 should be less than 2 (hence, 1 or 0) or very similar:
这意味着第2列和第3列中的NA之和应小于2(因此,1或0)或非常相似:
df[rowSums(is.na(df[c("age", "DOB")])) < 2L,]
And of course there's other options, like what @rawr provided in the comments.
当然还有其他选择,比如@rawr在评论中提供的内容。
And to better understand the subsetting, check this:
为了更好地理解子集,请检查以下内容:
rowSums(is.na(df[2:3]))
#[1] 0 0 0 2 0 0 1
rowSums(is.na(df[2:3])) < 2L
#[1] TRUE TRUE TRUE FALSE TRUE TRUE TRUE
#2
0
You were pretty close
你非常接近
df[!(is.na(df$age) & is.na(df$DOB)), ]
or
要么
df[!is.na(df$age) | !is.na(df$DOB), ]
#1
1
If you want to remove those rows where two columns (age and DOB) have more than 1 NA (which would mathematically mean that there could only be 2 NAs in such a case), you can do for example:
如果要删除两列(年龄和DOB)具有多于1个NA的行(这在数学上意味着在这种情况下只能有2个NA),您可以执行以下操作:
df[!is.na(df$age) | !is.na(df$DOB),]
which means that either both or one of the columns should be not NA, or
这意味着两个列或其中一个列不应该是NA,或者
df[rowSums(is.na(df[2:3])) < 2L,]
which means that the sum of NAs in columns 2 and 3 should be less than 2 (hence, 1 or 0) or very similar:
这意味着第2列和第3列中的NA之和应小于2(因此,1或0)或非常相似:
df[rowSums(is.na(df[c("age", "DOB")])) < 2L,]
And of course there's other options, like what @rawr provided in the comments.
当然还有其他选择,比如@rawr在评论中提供的内容。
And to better understand the subsetting, check this:
为了更好地理解子集,请检查以下内容:
rowSums(is.na(df[2:3]))
#[1] 0 0 0 2 0 0 1
rowSums(is.na(df[2:3])) < 2L
#[1] TRUE TRUE TRUE FALSE TRUE TRUE TRUE
#2
0
You were pretty close
你非常接近
df[!(is.na(df$age) & is.na(df$DOB)), ]
or
要么
df[!is.na(df$age) | !is.na(df$DOB), ]