英文:
r match and combine into a dataframe multiple vectors of different lengths
问题
结果：x y z[1,] "1" "1" NA[2,] "2" "2" NA[3,] NA "3" "3"[4,] NA "4" "4"[5,] NA "5" "5"[6,] NA "6" "6"[7,] "a" NA "a"[8,] "b" "b" NA
英文:
I am probably missing something very obvious, but I can't seem to find a way to do this. I would like to will merge multiple vectors (or dataframes?) of different lengths into a dataframe by matching values of vector elements with each other and putting them into same row positions, filling rows left empty with NAs. I have tried the solution from qpcR (cbind.na) but it doesn't produce expected outcome.
reproducible example:
x<-c("1","2","a","b")y<-c("1","2","3","4","5","6","b")z<-c("3","4","5","6","a")
expected output:
x y z[1,] 1 1 NA[2,] 2 2 NA[3,] NA 3 3[4,] NA 4 4[5,] NA 5 5[6,] NA 6 6[7,] a NA a[8,] b b NA
答案1
得分: 2
以下是翻译好的内容：
这是一个笨拙但有效的解决方案。请注意，行顺序与您的要求不匹配：
x <- c("1", "2", "a", "b")y <- c("1", "2", "3", "4", "5", "6", "b")z <- c("3", "4", "5", "6", "a")tot <- unique(c(x, y, z))# 给你一个包含所有向量中所有唯一值的列表df <- data.frame(x = rep(NA, times = length(tot)),y = NA,z = NA)# 准备一个包含所有NA值的数据框df$x[tot %in% x] <- tot[tot %in% x]df$y[tot %in% y] <- tot[tot %in% y]df$z[tot %in% z] <- tot[tot %in% z]# 如果在“父”向量中存在匹配值，则填充NA值。
结果：
> dfx y z1 1 1 <NA>2 2 2 <NA>3 a <NA> a4 b b <NA>5 <NA> 3 36 <NA> 4 47 <NA> 5 58 <NA> 6 6
英文:
Here's a clumsy but working solution. Note the row order does not match your request though:
x<-c("1","2","a","b")y<-c("1","2","3","4","5","6","b")z<-c("3","4","5","6","a")tot <- unique(c(x,y,z))# gives you a list of all unique values across all your vectorsdf <- data.frame(x = rep(NA, times = length(tot)),y = NA,z = NA)# prepare a data frame with all NAsdf$x[tot %in% x] <- tot[tot %in% x]df$y[tot %in% y] <- tot[tot %in% y]df$z[tot %in% z] <- tot[tot %in% z]# fills in the NAs with the matching value if present in the 'parent' vector.
Gives:
> dfx y z1 1 1 <NA>2 2 2 <NA>3 a <NA> a4 b b <NA>5 <NA> 3 36 <NA> 4 47 <NA> 5 58 <NA> 6 6
答案2
得分: 2
以下是翻译好的部分：
l <- list(x = x, y = y, z = z)dat <- data.frame(unique_vals = sort(unique(unlist(l))))dat[names(l)] <- lapply(l, \(x) {x[match(dat$unique_vals, x)]})# unique_vals x y z# 1 1 1 1 <NA># 2 2 2 2 <NA># 3 3 <NA> 3 3# 4 4 <NA> 4 4# 5 5 <NA> 5 5# 6 6 <NA> 6 6# 7 a a <NA> a# 8 b b b <NA>
我保留了unique_vals列以便清楚地了解操作，但你可能想要将其删除。
英文:
You could try this. It is similar to the answer by Paul Stafford Allen in that it starts with the unique values. I've put the vectors in a list to allow for easy iteration, so it is straightforward to extend to more columns.
l <- list(x = x, y = y, z = z)dat <- data.frame(unique_vals = sort(unique(unlist(l))))dat[names(l)] <- lapply(l, \(x) {x[match(dat$unique_vals, x)]})# unique_vals x y z# 1 1 1 1 <NA># 2 2 2 2 <NA># 3 3 <NA> 3 3# 4 4 <NA> 4 4# 5 5 <NA> 5 5# 6 6 <NA> 6 6# 7 a a <NA> a# 8 b b b <NA>
I kept the unique_vals column so it's clear what's going on but you may want to remove it.
答案3
得分: 1
你可以在Reduce中使用merge，并按新集合的行名称进行匹配。
l <- lapply(list(x=x, y=y, z=z), \(a) setNames(a, make.unique(a)))setNames(Reduce(\(a, b) {. <- merge(a, b, by=0, all=TRUE)`row.names<-`(.[-1], .[,1])}, l), names(l))# x y z#1 1 1 <NA>#2 2 2 <NA>#3 <NA> 3 3#4 <NA> 4 4#5 <NA> 5 5#6 <NA> 6 6#a a <NA> a#b b b <NA>
这也适用于向量中一个值出现多次的情况。
x<-c("1","1","2","a","b")y<-c("1","2","3","4","5","6","b")z<-c("3","4","5","6","a")l <- lapply(list(x=x, y=y, z=z), \(a) setNames(a, make.unique(a)))setNames(Reduce(\(a, b) {. <- merge(a, b, by=0, all=TRUE)`row.names<-`(.[-1], .[,1])}, l), names(l))# x y z#1 1 1 <NA>#1.1 1 <NA> <NA>#2 2 2 <NA>#3 <NA> 3 3#4 <NA> 4 4#5 <NA> 5 5#6 <NA> 6 6#a a <NA> a#b b b <NA>
或者使用match。
x <- c("1","1","2","a","b")y <- c("1","2","3","4","5","6","b")z <- c("3","4","5","6","a")l <- list(x=x, y=y, z=z)u <- lapply(l, make.unique)k <- unique(unlist(u))mapply(\(l, u) l[match(k, u)], l, u)# x y z# [1,] "1" "1" NA# [2,] "1" NA NA# [3,] "2" "2" NA# [4,] "a" NA "a"# [5,] "b" "b" NA# [6,] NA "3" "3"# [7,] NA "4" "4"# [8,] NA "5" "5"# [9,] NA "6" "6"
英文:
You can use merge in Reduce and match by the new set row.names.
l <- lapply(list(x=x, y=y, z=z), \(a) setNames(a, make.unique(a)))setNames(Reduce(\(a, b) {. <- merge(a, b, by=0, all=TRUE)`row.names<-`(.[-1], .[,1])}, l), names(l))# x y z#1 1 1 <NA>#2 2 2 <NA>#3 <NA> 3 3#4 <NA> 4 4#5 <NA> 5 5#6 <NA> 6 6#a a <NA> a#b b b <NA>
This will also work in case a value is more than one time present in a vector.
x<-c("1","1","2","a","b")y<-c("1","2","3","4","5","6","b")z<-c("3","4","5","6","a")l <- lapply(list(x=x, y=y, z=z), \(a) setNames(a, make.unique(a)))setNames(Reduce(\(a, b) {. <- merge(a, b, by=0, all=TRUE)`row.names<-`(.[-1], .[,1])}, l), names(l))# x y z#1 1 1 <NA>#1.1 1 <NA> <NA>#2 2 2 <NA>#3 <NA> 3 3#4 <NA> 4 4#5 <NA> 5 5#6 <NA> 6 6#a a <NA> a#b b b <NA>
Or using match.
x <- c("1","1","2","a","b")y <- c("1","2","3","4","5","6","b")z <- c("3","4","5","6","a")l <- list(x=x, y=y, z=z)u <- lapply(l, make.unique)k <- unique(unlist(u))mapply(\(l, u) l[match(k, u)], l, u)# x y z# [1,] "1" "1" NA# [2,] "1" NA NA# [3,] "2" "2" NA# [4,] "a" NA "a"# [5,] "b" "b" NA# [6,] NA "3" "3"# [7,] NA "4" "4"# [8,] NA "5" "5"# [9,] NA "6" "6"
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