R 将自定义函数应用于每行只使用参数的第一个值
我正在尝试使用以下数据集将R 将自定义函数应用于每行只使用参数的第一个值,r,apply,na,missing-data,R,Apply,Na,Missing Data,我正在尝试使用以下数据集将NA值重新编码为0: set.seed(1) df <- data.frame( id = c(1:10), trials = sample(1:3, 10, replace = T), t1 = c(sample(c(1:9, NA), 10)), t2 = c(sample(c(1:7, rep(NA, 3)), 10)), t3 = c(sample(c(1:5, rep(NA, 5)), 10)) ) 这适用于单个向量。当我尝试使
NA
值重新编码为0
:
set.seed(1)
df <- data.frame(
id = c(1:10),
trials = sample(1:3, 10, replace = T),
t1 = c(sample(c(1:9, NA), 10)),
t2 = c(sample(c(1:7, rep(NA, 3)), 10)),
t3 = c(sample(c(1:5, rep(NA, 5)), 10))
)
这适用于单个向量。当我尝试使用apply()
将同一函数应用于数据帧时,尽管:
apply(df, 1, replace0, num.sun = df$trials)
我收到一条警告说:
In 1:(num.sun + 2) :
numerical expression has 10 elements: only the first used
结果是,apply()
只对每一行使用trials
列中的第一个值,而不是让num.sun
的值根据trials
中的值更改每一行。如何应用该函数,使num.sun
参数根据df$trials
的值变化
谢谢
编辑:正如一些人评论的那样,原始示例数据有一些非NA分数,根据试验列,这些分数没有意义。以下是更正后的数据集:
df <- data.frame(
id = c(1:5),
trials = c(rep(1, 2), rep(2, 1), rep(3, 2)),
t1 = c(NA, 7, NA, 6, NA),
t2 = c(NA, NA, 3, 7, 12),
t3 = c(NA, NA, NA, 4, NA)
)
df以下是一种方法:
x <- is.na(df)
df[x & t(apply(x, 1, cumsum)) > 3 - df$trials] <- 0
x3-df$trials]df
id试验t1 t2 t3
1 1 3 NA 5
2不适用
3 3 2 6 6 4
4 4 3 0 1 2
515NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 1
9 9 2 1 3 NA
10 10 1 9 4 3
>x df[x&t(应用(x,1,cumsum))>3-df$trials]df
id试验t1 t2 t3
1 1 3 NA 5
2不适用
3 3 2 6 6 4
4 4 3 0 1 2
515NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 1
9 9 2 1 3 NA
10 10 1 9 4 3
注意:第1/3/10行是有问题的,因为非NA值比试验值多。这里我只是使用双子集x[paste0('t',x['trials'])]]
重写了您的函数,这克服了第6行的其他两个解决方案中的问题
replace0 <- function(x){
#browser()
x_na <- x[paste0('t',x['trials'])]
if(is.na(x_na)){x[paste0('t',x['trials'])] <- 0}
return(x)
}
t(apply(df, 1, replace0))
id trials t1 t2 t3
[1,] 1 1 3 NA 5
[2,] 2 2 2 2 NA
[3,] 3 2 6 6 4
[4,] 4 3 NA 1 2
[5,] 5 1 5 NA NA
[6,] 6 3 7 NA 0
[7,] 7 3 8 7 0
[8,] 8 2 4 5 1
[9,] 9 2 1 3 NA
[10,] 10 1 9 4 3
replace0另一种方法:
# create an index of the NA values
w <- which(is.na(df), arr.ind = TRUE)
# create an index with the max column by row where an NA is allowed to be replaced by a zero
m <- matrix(c(1:nrow(df), (df$trials + 2)), ncol = 2)
# subset 'w' such that only the NA's which fall in the scope of 'm' remain
i <- w[w[,2] <= m[,2][match(w[,1], m[,1])],]
# use 'i' to replace the allowed NA's with a zero
df[i] <- 0
您可以轻松地将其包装到函数中:
replace0 <- function(x, num.sun) {
x[which(is.na(x[1:(num.sun + 2)]))] <- 0
return(x)
}
replace.NA.with.0 <- function(df) {
w <- which(is.na(df), arr.ind = TRUE)
m <- matrix(c(1:nrow(df), (df$trials + 2)), ncol = 2)
i <- w[w[,2] <= m[,2][match(w[,1], m[,1])],]
df[i] <- 0
return(df)
}
现在,使用replace.with.NA.or.0(df)
生成以下结果:
这里有一个tidyverse
方法,请注意,它不会提供与其他解决方案相同的输出
您的示例数据显示了“没有发生”的试验结果,我假设您的真实数据没有
library(tidyverse)
df %>%
nest(matches("^t\\d")) %>%
mutate(data = map2(data,trials,~mutate_all(.,replace_na,0) %>% select(.,1:.y))) %>%
unnest
# id trials t1 t2 t3
# 1 1 1 3 NA NA
# 2 2 2 2 2 NA
# 3 3 2 6 6 NA
# 4 4 3 0 1 2
# 5 5 1 5 NA NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
# 8 8 2 4 5 NA
# 9 9 2 1 3 NA
# 10 10 1 9 NA NA
使用更常用的收集
策略,这将是:
df %>%
gather(k,v,matches("^t\\d")) %>%
arrange(id) %>%
group_by(id) %>%
slice(1:first(trials)) %>%
mutate_at("v",~replace(.,is.na(.),0)) %>%
spread(k,v)
# # A tibble: 10 x 5
# # Groups: id [10]
# id trials t1 t2 t3
# <int> <int> <dbl> <dbl> <dbl>
# 1 1 1 3 NA NA
# 2 2 2 2 2 NA
# 3 3 2 6 6 NA
# 4 4 3 0 1 2
# 5 5 1 5 NA NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
# 8 8 2 4 5 NA
# 9 9 2 1 3 NA
# 10 10 1 9 NA NA
df%>%
聚集(k,v,匹配项(^t\\d))%>%
排列(id)%>%
分组依据(id)%>%
切片(1:首次(试验))%>%
在(“v”,~replace(,is.na(.),0))%>%
扩散(k,v)
##A tible:10 x 5
##组:id[10]
#id试验t1 t2 t3
#
#1 1 3 NA NA
#2不适用
#3 3 2 6 NA
# 4 4 3 0 1 2
#515NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
#8245NA
#9 9 2 1 3 NA
#101019NA
第3行和第10行也是problematic@Moody_Mudskipper,谢谢你的来信。我编辑了答案以反映这一点。
replace.with.NA.or.0 <- function(df) {
w <- which(is.na(df), arr.ind = TRUE)
df[w] <- 0
v <- tapply(m[,2], m[,1], FUN = function(x) tail(x:5,-1))
ina <- matrix(as.integer(unlist(stack(v)[2:1])), ncol = 2)
df[ina] <- NA
return(df)
}
id trials t1 t2 t3
1 1 1 3 NA NA
2 2 2 2 2 NA
3 3 2 6 6 NA
4 4 3 0 1 2
5 5 1 5 NA NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 NA
9 9 2 1 3 NA
10 10 1 9 NA NA
library(tidyverse)
df %>%
nest(matches("^t\\d")) %>%
mutate(data = map2(data,trials,~mutate_all(.,replace_na,0) %>% select(.,1:.y))) %>%
unnest
# id trials t1 t2 t3
# 1 1 1 3 NA NA
# 2 2 2 2 2 NA
# 3 3 2 6 6 NA
# 4 4 3 0 1 2
# 5 5 1 5 NA NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
# 8 8 2 4 5 NA
# 9 9 2 1 3 NA
# 10 10 1 9 NA NA
df %>%
gather(k,v,matches("^t\\d")) %>%
arrange(id) %>%
group_by(id) %>%
slice(1:first(trials)) %>%
mutate_at("v",~replace(.,is.na(.),0)) %>%
spread(k,v)
# # A tibble: 10 x 5
# # Groups: id [10]
# id trials t1 t2 t3
# <int> <int> <dbl> <dbl> <dbl>
# 1 1 1 3 NA NA
# 2 2 2 2 2 NA
# 3 3 2 6 6 NA
# 4 4 3 0 1 2
# 5 5 1 5 NA NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
# 8 8 2 4 5 NA
# 9 9 2 1 3 NA
# 10 10 1 9 NA NA