英文:
How to fill NA rows by conditions from columns in R
问题
这是一个示例:
df <- data.frame(v1=rep(1:2, 4),
v2=rep(c("a", "b"), each=4),
v3=paste0(rep(1:2, each=4), rep(c("m", "n", "o", "p"), each=2)),
v4=c(1,2, NA, NA, 3,4, NA,NA),
v5=c(5,6, NA, NA, 7,8, NA,NA),
v6=c(9,10, NA, NA, 11,12, NA,NA))
df
这是我想要的,如果列v1
+v2
+v3
相同,忽略v3
的最后一个字母,那么将从不是NA
的行中填充NAs
。在这种情况下,由于1a1相同(忽略m),所以行3的NA应该由行1填充。所以期望的输出是:
v1 v2 v3 v4 v5 v6
1 1 a 1m 1 5 9
2 2 a 1m 2 6 10
3 1 a 1n 1 5 9
4 2 a 1n 2 6 10
5 1 b 2o 3 7 11
6 2 b 2o 4 8 12
7 1 b 2p 3 7 11
8 2 b 2p 4 8 12
英文:
Here is an example:
df<-data.frame(v1=rep(1:2, 4),
v2=rep(c("a", "b"), each=4),
v3=paste0(rep(1:2, each=4), rep(c("m", "n", "o", "p"), each=2)),
v4=c(1,2, NA, NA, 3,4, NA,NA),
v5=c(5,6, NA, NA, 7,8, NA,NA),
v6=c(9,10, NA, NA, 11,12, NA,NA))
df
v1 v2 v3 v4 v5 v6
1 1 a 1m 1 5 9
2 2 a 1m 2 6 10
3 1 a 1n NA NA NA
4 2 a 1n NA NA NA
5 1 b 2o 3 7 11
6 2 b 2o 4 8 12
7 1 b 2p NA NA NA
8 2 b 2p NA NA NA
What I wanted is, if column v1
+v2
+v3
are same by ignore the last letter of v3
, fill the NAs
from the rows that are not NA
. In this case, row3's NA should be filled by row1 due to same 1a1 by ignoring m. So a desired output would be:
v1 v2 v3 v4 v5 v6
1 1 a 1m 1 5 9
2 2 a 1m 2 6 10
3 1 a 1n 1 5 9
4 2 a 1n 2 6 10
5 1 b 2o 3 7 11
6 2 b 2o 4 8 12
7 1 b 2p 3 7 11
8 2 b 2p 4 8 12
答案1
得分: 2
我不知道,但我认为这是生成您的结果的更简单方法
library(tidyverse)
df %>%
group_by(v1,v2) %>%
fill(v4:v6)
# 添加v3逻辑
df %>%
mutate(v7 = v3 %>%
as.character() %>%
parse_number()) %>%
group_by(v1,v2,v7) %>%
fill(v4:v6) %>%
select(-v7)
英文:
I don't know but I think this is a simpler way of producing your results
library(tidyverse)
df %>%
group_by(v1,v2) %>%
fill(v4:v6)
Adding the v3 logic
df %>%
mutate(v7 = v3 %>% as.character() %>% parse_number()) %>%
group_by(v1,v2,v7) %>%
fill(v4:v6) %>%
select(-v7)
答案2
得分: 1
这是一个将 v3
重新编码为仅考虑数字部分的变量的解决方案。
library(dplyr)
library(stringr)
# 提取字符串 v3 中的数字部分
df$v7 <- str_extract(df$v3, "[[:digit:]]+")
df %>%
group_by(v1, v2, v7) %>%
fill(v4:v6)
英文:
Here is a solution that recodes v3
into a variable that only takes into account the numeric part.
library(dplyr)
library(stringr)
#Extract numeric part of the string in v3
df$v7<-str_extract(df$v3,"[[:digit:]]+")
df %>%
group_by(v1,v2,v7) %>%
fill(v4:v6)
答案3
得分: 0
使用zoo
中的na.locf
library(zoo)
library(data.table)
setDT(df)[, na.locf(.SD), .(v1, v2)]
# v1 v2 v3 v4 v5 v6
#1: 1 a 1m 1 5 9
#2: 1 a 1n 1 5 9
#3: 2 a 1m 2 6 10
#4: 2 a 1n 2 6 10
#5: 1 b 2o 3 7 11
#6: 1 b 2p 3 7 11
#7: 2 b 2o 4 8 12
#8: 2 b 2p 4 8 12
如果我们想要在'v3'中添加条件
setDT(df)[, names(df)[4:6] := na.locf(.SD), .(v1, v2, sub("\\D+", "", v3))][]
# v1 v2 v3 v4 v5 v6
#1: 1 a 1m 1 5 9
#2: 2 a 1m 2 6 10
#3: 1 a 1n 1 5 9
#4: 2 a 1n 2 6 10
#5: 1 b 2o 3 7 11
#6: 2 b 2o 4 8 12
#7: 1 b 2p 3 7 11
#8: 2 b 2p 4 8 12
英文:
Using na.locf
from zoo
library(zoo)
library(data.table)
setDT(df)[, na.locf(.SD),.(v1, v2)]
# v1 v2 v3 v4 v5 v6
#1: 1 a 1m 1 5 9
#2: 1 a 1n 1 5 9
#3: 2 a 1m 2 6 10
#4: 2 a 1n 2 6 10
#5: 1 b 2o 3 7 11
#6: 1 b 2p 3 7 11
#7: 2 b 2o 4 8 12
#8: 2 b 2p 4 8 12
If we want to add the condition in 'v3'
setDT(df)[, names(df)[4:6] := na.locf(.SD),.(v1, v2, sub("\\D+", "", v3))][]
# v1 v2 v3 v4 v5 v6
#1: 1 a 1m 1 5 9
#2: 2 a 1m 2 6 10
#3: 1 a 1n 1 5 9
#4: 2 a 1n 2 6 10
#5: 1 b 2o 3 7 11
#6: 2 b 2o 4 8 12
#7: 1 b 2p 3 7 11
#8: 2 b 2p 4 8 12
答案4
得分: 0
以下是使用 data.table
和 zoo
的解决方案,忽略了 v3
列的最后一个字母:
library(data.table)
setDT(df)[, match_cols := paste0(v1, v2, substr(v3, 1, nchar(as.character(v3)) - 1))][, id := .GRP, by = match_cols][, v4 := zoo::na.locf(v4, na.rm = F), by = id][, v5 := zoo::na.locf(v5, na.rm = F), by = id][, v6 := zoo::na.locf(v6, na.rm = F), by = id][ , c("match_cols", "id") := NULL]
df
结果如下:
v1 v2 v3 v4 v5 v6
1: 1 a 1m 1 5 9
2: 2 a 1m 2 6 10
3: 1 a 1n 1 5 9
4: 2 a 1n 2 6 10
5: 1 b 2o 3 7 11
6: 2 b 2o 4 8 12
7: 1 b 2p 3 7 11
8: 2 b 2p 4 8 12
英文:
Here's a solution using data.table
and zoo
which ignores v3
column's last letter:
library(data.table)
setDT(df)[, match_cols := paste0(v1, v2, substr(v3, 1, nchar(as.character(v3)) - 1))][, id := .GRP, by = match_cols][, v4 := zoo::na.locf(v4, na.rm = F), by = id][, v5 := zoo::na.locf(v5, na.rm = F), by = id][, v6 := zoo::na.locf(v6, na.rm = F), by = id][ , c("match_cols", "id") := NULL]
df
# v1 v2 v3 v4 v5 v6
#1: 1 a 1m 1 5 9
#2: 2 a 1m 2 6 10
#3: 1 a 1n 1 5 9
#4: 2 a 1n 2 6 10
#5: 1 b 2o 3 7 11
#6: 2 b 2o 4 8 12
#7: 1 b 2p 3 7 11
#8: 2 b 2p 4 8 12
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