英文:
Replace NA values in dataframe with variables in the next column (R)
问题
我是新手，还在努力学习R语言，但在其他帖子中找不到我要找的答案。
我有一个数据集，简化起见，有5列。第1、2和4列总是有值的，但在某些行中，第3列没有值。以下是一个示例：
当前数据集：
A B C D E
1 1 2 3
1 2 NA 4 5
1 2 3 4
1 3 NA 9 7
1 2 NA 5 6
我想要将NA值替换为第D列的值，然后将第E列的值移到D列，以此类推。
期望的输出：
A B C D E
1 1 2 3 NA
1 2 4 5 NA
1 2 3 4 NA
1 3 9 7 NA
1 2 5 6 NA
我复制了不同的Stack Overflow帖子中的代码，但都没有达到我想要的效果。
na.omit会删除行。非常感谢任何帮助。
英文:
I am new still trying to learn R and I could not find the answers I am looking for in any other thread.
I have a dataset with (for simplicity) 5 columns. Columns 1,2, and4 always have values, but in some rows column 3 doesn't. Below is an example:
Current
A B C D E
1 1 2 3
1 2 NA 4 5
1 2 3 4
1 3 NA 9 7
1 2 NA 5 6
I want to make it so that the NA's are replaced by the value in column D, and then the value in col E is shifted to D, etc.
Desired output:
A B C D E
1 1 2 3 NA
1 2 4 5 NA
1 2 3 4 NA
1 3 9 7 NA
1 2 5 6 NA
I copied what was on different Stack overflow threads and none achieved what I wanted.
na.omit gets rid of the row. Any help is greatly appreciated.
答案1
得分: 1
以下是翻译好的内容：
## 数据
data <- structure(list(A = c(1L, 1L, 1L, 1L, 1L), B = c(1L, 2L, 2L, 3L,
2L), C = c(2L, NA, 3L, NA, NA), D = c(3L, 4L, 4L, 9L, 5L), E = c(NA,
5L, NA, 7L, 6L)), class = "data.frame", row.names = c(NA, -5L
))
## 代码
library(dplyr)
data %>%
mutate(
aux = C,
C = if_else(is.na(aux), D, C),
D = if_else(is.na(aux), E, D),
E = NA
) %>%
select(-aux)
## 输出
A B C D E
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
希望这有帮助！
英文:
Data
data <- structure(list(A = c(1L, 1L, 1L, 1L, 1L), B = c(1L, 2L, 2L, 3L,
2L), C = c(2L, NA, 3L, NA, NA), D = c(3L, 4L, 4L, 9L, 5L), E = c(NA,
5L, NA, 7L, 6L)), class = "data.frame", row.names = c(NA, -5L
))
Code
library(dplyr)
data %>%
mutate(
aux = C,
C = if_else(is.na(aux),D,C),
D = if_else(is.na(aux),E,D),
E = NA
) %>%
select(-aux)
Output
A B C D E
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
答案2
得分: 1
替换操作一次完成：
dat[is.na(dat$C), c("C","D","E")] <- c(dat[is.na(dat$C), c("D","E")], NA)
dat
# A B C D E
#1 1 1 2 3 NA
#2 1 2 4 5 NA
#3 1 2 3 4 NA
#4 1 3 9 7 NA
#5 1 2 5 6 NA
其中 dat 是：
dat <- read.table(text="A B C D E
1 1 2 3
1 2 NA 4 5
1 2 3 4
1 3 NA 9 7
1 2 NA 5 6", fill=TRUE, header=TRUE)
英文:
Replacement operation all in one go:
dat[is.na(dat$C), c("C","D","E")] <- c(dat[is.na(dat$C), c("D","E")], NA)
dat
# A B C D E
#1 1 1 2 3 NA
#2 1 2 4 5 NA
#3 1 2 3 4 NA
#4 1 3 9 7 NA
#5 1 2 5 6 NA
Where dat was:
dat <- read.table(text="A B C D E
1 1 2 3
1 2 NA 4 5
1 2 3 4
1 3 NA 9 7
1 2 NA 5 6", fill=TRUE, header=TRUE)
答案3
得分: 1
使用 shift_row_values
library(hacksaw)
shift_row_values(df1)
A B C D E
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
数据
df1 <- structure(list(A = c(1L, 1L, 1L, 1L, 1L), B = c(1L, 2L, 2L, 3L,
2L), C = c(2L, NA, 3L, NA, NA), D = c(3L, 4L, 4L, 9L, 5L), E = c(NA,
5L, NA, 7L, 6L)), class = "data.frame", row.names = c(NA, -5L
))
英文:
Using shift_row_values
library(hacksaw)
shift_row_values(df1)
A B C D E
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
data
df1 <- structure(list(A = c(1L, 1L, 1L, 1L, 1L), B = c(1L, 2L, 2L, 3L,
2L), C = c(2L, NA, 3L, NA, NA), D = c(3L, 4L, 4L, 9L, 5L), E = c(NA,
5L, NA, 7L, 6L)), class = "data.frame", row.names = c(NA, -5L
))
答案4
得分: 0
以下是您要的翻译内容：
A base R universal approach using order without prior knowledge of NA positions.
setNames(data.frame(t(apply(data, 1, function(x)
x[order(is.na(x))]))), colnames(data))
A B C D E
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
使用 dplyr
library(dplyr)
t(data) %>%
data.frame() %>%
mutate(across(everything(), ~ .x[order(is.na(.x))])) %>%
t() %>%
as_tibble()
# A tibble: 5 × 5
A B C D E
<int> <int> <int> <int> <int>
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
数据
data <- structure(list(A = c(1L, 1L, 1L, 1L, 1L), B = c(1L, 2L, 2L, 3L,
2L), C = c(2L, NA, 3L, NA, NA), D = c(3L, 4L, 4L, 9L, 5L), E = c(NA,
5L, NA, 7L, 6L)), class = "data.frame", row.names = c(NA, -5L
))
英文:
A base R universal approach using order without prior knowledge of NA positions.
setNames(data.frame(t(apply(data, 1, function(x)
x[order(is.na(x))]))), colnames(data))
A B C D E
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
Using dplyr
library(dplyr)
t(data) %>%
data.frame() %>%
mutate(across(everything(), ~ .x[order(is.na(.x))])) %>%
t() %>%
as_tibble()
# A tibble: 5 × 5
A B C D E
<int> <int> <int> <int> <int>
1 1 1 2 3 NA
2 1 2 4 5 NA
3 1 2 3 4 NA
4 1 3 9 7 NA
5 1 2 5 6 NA
Data
data <- structure(list(A = c(1L, 1L, 1L, 1L, 1L), B = c(1L, 2L, 2L, 3L,
2L), C = c(2L, NA, 3L, NA, NA), D = c(3L, 4L, 4L, 9L, 5L), E = c(NA,
5L, NA, 7L, 6L)), class = "data.frame", row.names = c(NA, -5L
))
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