[英]Replace NA with values in another row of same column for each group in r - values not unique within group
I have a question very similar to a previous one but I am unable to generalize it to my case.我有一个与前一个问题非常相似的问题,但我无法将其概括为我的案例。
I have data that looks sort of like this我有看起来像这样的数据
Within each ID, I have several Vis rows.在每个 ID 中,我有几个 Vis 行。 The ones of interest to me are only a and b .我感兴趣的只有a和b 。 The data is such that for each column in the data (V1...V7), if a is present, b is present and for all values of a , b is missing and vice versa.数据是这样的,在数据(V1 ... V7)的每一列中,如果存在时,b的存在和用于所有的a值,b为丢失,并且反之亦然。 I would like to combine Vis's a and b for each ID group such that I have a single row (either a or b or even a new one, it doesn't really matter) without any missing data for any of the columns.我想为每个 ID 组组合 Vis 的a和b ,这样我就有一行(a 或 b 或什至是一个新行,这并不重要)而没有任何列的任何缺失数据。
Based on the image showed, may be this helps.根据显示的图像,这可能有帮助。 Here I am using actual NAs with only a couple of V columns.在这里,我使用只有几个 V 列的实际 NA。
We create a numeric index for column names that start with 'V' followed by numbers ('nm1').我们为以“V”开头、后跟数字 (“nm1”) 的列名创建数字索引。 Convert the 'data.frame' to 'data.table' ( setDT(df1)
), grouped by 'ID', we use Map
, loop over the columns specified by the index 'nm1' ( SD[, nm1, with=FALSE]
) and the 'Vis' column, replace
the 'V' column elements where the 'Vis' is either 'a' or 'b' by the non-NA element ( na.omit(x[..
), and assign the output to the numeric index.将 'data.frame' 转换为 'data.table' ( setDT(df1)
),按 'ID' 分组,我们使用Map
,循环遍历由索引 'nm1' 指定的列( SD[, nm1, with=FALSE]
) 和 'Vis' 列, replace
'Vis' 为 'a' 或 'b' 的 'V' 列元素replace
为非 NA 元素 ( na.omit(x[..
),并分配输出到数字索引。
library(data.table)
nm1 <- grep('V\\d+',colnames(df1))
setDT(df1)[, (nm1):= Map(function(x,y)
replace(x, which(y %in% c('a', 'b')), na.omit(x[y %in% c('a', 'b')])),
.SD[,-1, with=FALSE], list(.SD[[1]])), ID]
We change the 'b' values to 'a'我们将“b”值更改为“a”
df1[Vis=='b', Vis := 'a']
and get the unique
rows并获得unique
行
unique(df1)
# ID Vis V1 V2
#1: 2 a 1 2
#2: 2 c 4 5
#3: 3 a 3 4
#4: 4 a 2 3
#5: 4 c 3 4
#6: 4 d 1 1
df1 <- data.frame(ID= rep(c(2,3,4), c(3,2,4)), Vis=c('a', 'b', 'c', 'a',
'b', 'a', 'b', 'c', 'd'), V1= c(1, NA, 4, 3, NA, NA, 2, 3, 1),
V2= c(NA, 2, 5, 4, NA, 3, NA, 4, 1), stringsAsFactors=FALSE)
Just sum the values you need while removing NAs.只需在删除 NA 时对您需要的值求和即可。 There are more vectorized ways to do this, but the for loop is a bit clearer.有更多的矢量化方法可以做到这一点,但 for 循环更清晰一些。
for(I in unique(df1$ID)) {
df_sub <- subset(df1, df1$ID==I & df1$Vis %in% c("a", "b"))
df1 <- subset(df1, df1$ID != I)
new_row <- apply(df_sub[, -1:-2], 2, sum, na.rm=TRUE)
df1 <- rbind(df1, c(ID=I, new_row))
}
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