英文:
Fill across a data frame based on a value in a column in R
问题
我想要填充在y中指定的列数中m0中的值。期望的结果是：
| y | m0 | m1 | m2 | mn |
|---|---|---|---|---|
| 1 | 5 | 5 | NA | NA |
| 2 | 15 | 15 | 15 | NA |
| 3 | 25 | 25 | 25 | 25 |
感谢任何指导！
英文:
I have a data frame as below:
| y | m0 | m1 | m2 | mn |
|---|---|---|---|---|
| 1 | 5 | NA | NA | NA |
| 2 | 15 | NA | NA | NA |
| 3 | 25 | NA | NA | NA |
I would like to fill the value in m0 across the number of columns specified in y. Desired result is:
| y | m0 | m1 | m2 | mn |
|---|---|---|---|---|
| 1 | 5 | 5 | NA | NA |
| 2 | 15 | 15 | 15 | NA |
| 3 | 25 | 25 | 25 | 25 |
Appreciate any guidance!
答案1
得分: 1
你可以首先将你的数据转换成“长”格式，然后在NA的计数在y内时，用first值替换NA。
library(tidyverse)df %>%pivot_longer(-y) %>%mutate(value = ifelse(is.na(value) & cumsum(is.na(value)) <= y, first(value), value),.by = y) %>%pivot_wider()#> # A tibble: 3 × 5#> y m0 m1 m2 mn#> <int> <int> <int> <int> <int>#> 1 1 5 5 NA NA#> 2 2 15 15 15 NA#> 3 3 25 25 25 25
请注意，这是R语言中的代码示例，用于将数据从宽格式转换为长格式，然后在特定条件下替换缺失值。
英文:
You can first transform your data to a "long" format, then replace NA with the first value when the count of NA is within y.
library(tidyverse)df %>%pivot_longer(-y) %>%mutate(value = ifelse(is.na(value) & cumsum(is.na(value)) <= y, first(value), value),.by = y) %>%pivot_wider()#> # A tibble: 3 × 5#> y m0 m1 m2 mn#> <int> <int> <int> <int> <int>#> 1 1 5 5 NA NA#> 2 2 15 15 15 NA#> 3 3 25 25 25 25
答案2
得分: 1
A 基本 R 方法使用 sapply。 colbeg 定义了第一个需要替换 NA 的列。可以硬编码，但我怀疑它应该可调整为真实数据。
colbeg <- 3data.frame(t(sapply(seq_along(df$y), \(x){df[x,colbeg:(colbeg - 1 + df$y[x])] <- df$m0[x]; df[x,]})))y m0 m1 m2 mn1 1 5 5 NA NA2 2 15 15 15 NA3 3 25 25 25 25
英文:
A base R approach using sapply. colbeg defines the first column where NA has to be replaced. Can be hard coded but I suspect that it should be adjustable for the real data.
colbeg <- 3data.frame(t(sapply(seq_along(df$y), \(x){df[x,colbeg:(colbeg - 1 + df$y[x])] <- df$m0[x]; df[x,]})))y m0 m1 m2 mn1 1 5 5 NA NA2 2 15 15 15 NA3 3 25 25 25 25
答案3
得分: 0
Base R:```rlapply(setNames(0:2, paste0("m", 0:2 + 1)), function(z) if (z < 1) quux$m0 else c(rep(NA, z), tail(quux$m0, n = -z))) |as.data.frame()# m1 m2 m3# 1 5 NA NA# 2 15 15 NA# 3 25 25 25
You can either replace the columns or cbind them.
quux[,0:2 + 3] <- lapply(setNames(0:2, paste0("m", 0:2 + 1)), function(z) if (z < 1) quux$m0 else c(rep(NA, z), tail(quux$m0, n = -z))) |as.data.frame()quux <- cbind(quux[,1:2],lapply(setNames(0:2, paste0("m", 0:2 + 1)), function(z) if (z < 1) quux$m0 else c(rep(NA, z), tail(quux$m0, n = -z))))
(I'm consistently using 0:2 and offsets to show how things would be changed for an arbitrary number of columns.)
<details><summary>英文:</summary>Base R:```rlapply(setNames(0:2, paste0("m", 0:2 + 1)), function(z) if (z < 1) quux$m0 else c(rep(NA, z), tail(quux$m0, n = -z))) |>as.data.frame()# m1 m2 m3# 1 5 NA NA# 2 15 15 NA# 3 25 25 25
You can either replace the columns or cbind them.
quux[,0:2 + 3] <- lapply(setNames(0:2, paste0("m", 0:2 + 1)), function(z) if (z < 1) quux$m0 else c(rep(NA, z), tail(quux$m0, n = -z))) |>as.data.frame()quux <- cbind(quux[,1:2],lapply(setNames(0:2, paste0("m", 0:2 + 1)), function(z) if (z < 1) quux$m0 else c(rep(NA, z), tail(quux$m0, n = -z))))
(I'm consistently using 0:2 and offsets to show how things would be changed for an arbitrary number of columns.)
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