如何查看第一个NA在数据框中的一行中出现的位置(在哪一列中(？
我正在研究参与者在一个长达 10 个步骤的过程中放弃的点。
每个步骤都通过相应的列标识，这意味着总共 10 列。
我可以判断某人是否完成了任何一个步骤的方法是，如果我在列中看到一个日期时间值，该值指示他们完成该步骤的时刻。
如果他们尚未完成该步骤，它将显示NA，以下列也是如此。
例如，如果我在第 5 列中看到特定行的NA，那么我知道该特定用户没有继续执行步骤 4，因为其余列也将显示NA。
这个想法是参与者完成所有 10 个步骤，这意味着他们已经完全完成了该过程。
我希望能够确定最常见的下车点。
我的数据集有 2,000 行深 - 如何快速检查和/或识别？
示例数据：
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如果你想将列的名称添加到你的 data.frame 中，这样的事情就可以了：
创建一些数据，MTCARS的前10条记录。 将您在NA的情况复制到最后。
df <- mtcars[1:10, ]
df[3, 3:11] <- NA
df[6, 5:11] <- NA
df$dropofpoint <- apply(df, 1, function(x) names(which(is.na(x)))[1])
head(df)
mpg cyl disp hp drat wt qsec vs am gear carb dropofpoint
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 <NA>
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 <NA>
Datsun 710 22.8 4 NA NA NA NA NA NA NA NA NA disp
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 <NA>
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 <NA>
Valiant 18.1 6 225 105 NA NA NA NA NA NA NA drat
如果你想要一个快速的解决方案，我会使用max.col
res <- max.col(is.na(df), ties = "first")
尽管即使特定行中根本没有NA，max.col也会返回1。因此，您可以添加以下行来处理这些特定情况
if(any(res == 1)) is.na(res) <- (res == 1) & !is.na(df[[1]])
这会将这些事例转换为NA- 这意味着找不到该行的列索引
在下面的代码中，我假设您希望返回一个特殊值，以防一行中没有NA。
set.seed(5239) # Make the results reproducible
dat <- matrix(1:40, 4)
dat[sample(40, 5)] <- NA
dat <- as.data.frame(dat) # Not strictly needed
apply(dat, 1, function(x) {
w <- which(is.na(x))
if(length(w) > 0) min(w) else Inf
})
#[1] 4 6 Inf 2
这里的特殊价值是Inf.例如，您可以将其更改为length(x)或您选择的其他值。
下面是一个示例：
## some fake data
Data <- matrix(c(0,0,0,NA,0,NA,NA,NA,0,0,NA,NA), nrow = 3, byrow = TRUE)
## which ones are the first NA's per row
## Edited to avoid the warning message
apply(Data,1, function(fo) ifelse(any(is.na(fo)),min(which(is.na(fo))), NA))
考虑一个数据框，它有 7 行和 3 列，第一个 na 在第 4 行第 2 列。
df <- read.table(text = "rowname value test
A 3 8
B 1 3
C 2 8
D NA 5
E 2 3
F NA 4
G 6 9", header = TRUE)
然后可以使用哪个和 is.na 找到数据帧中第一个 na 值的位置。 例如：第一个 na 位于索引 1 处。可以遍历每个索引以按顺序获取后续 NAS。
pos_first_na <- which(is.na(df))[1]
要查找确切的行和列，请执行以下操作：
col_pos <- pos_first_na %% ncol(df)
row_pos <- pos_first_na %% nrow(df)
df[row_pos,col_pos]
这给出了：
> row_pos
[1] 4
> col_pos
[1] 2
> df[row_pos,col_pos]
[1] NA
请考虑以下解决方案，将x替换为您的数据：
# Loop through each row of x
apply(X = x, MARGIN = 1, FUN = function(row) {
# Return smallest index of NA values
min(which(is.na(row)))
})