jor*_*ran 4 r dplyr tidyr rlang
通常我需要spread多个值列,就像在这个问题中一样.但我经常这样做,我希望能够编写一个这样做的功能.
例如,给定数据:
set.seed(42)
dat <- data_frame(id = rep(1:2,each = 2),
grp = rep(letters[1:2],times = 2),
avg = rnorm(4),
sd = runif(4))
> dat
# A tibble: 4 x 4
id grp avg sd
<int> <chr> <dbl> <dbl>
1 1 a 1.3709584 0.6569923
2 1 b -0.5646982 0.7050648
3 2 a 0.3631284 0.4577418
4 2 b 0.6328626 0.7191123
我想创建一个返回类似的函数:
# A tibble: 2 x 5
id a_avg b_avg a_sd b_sd
<int> <dbl> <dbl> <dbl> <dbl>
1 1 1.3709584 -0.5646982 0.6569923 0.7050648
2 2 0.3631284 0.6328626 0.4577418 0.7191123
我怎样才能做到这一点？
我们将回到与之相关的问题中提供的答案,但是暂时让我们从一个更天真的方法开始.
一个想法是分别对spread每个值列,然后加入结果,即
library(dplyr)
library(tidyr)
library(tibble)
dat_avg <- dat %>%
select(-sd) %>%
spread(key = grp,value = avg) %>%
rename(a_avg = a,
b_avg = b)
dat_sd <- dat %>%
select(-avg) %>%
spread(key = grp,value = sd) %>%
rename(a_sd = a,
b_sd = b)
> full_join(dat_avg,
dat_sd,
by = 'id')
# A tibble: 2 x 5
id a_avg b_avg a_sd b_sd
<int> <dbl> <dbl> <dbl> <dbl>
1 1 1.3709584 -0.5646982 0.6569923 0.7050648
2 2 0.3631284 0.6328626 0.4577418 0.7191123
(我使用了一个full_join以防万一我们遇到并非所有连接列的组合出现在所有组合中的情况.)
让我们从一个类似spread但允许您将列key和value列作为字符传递的函数开始:
spread_chr <- function(data, key_col, value_cols, fill = NA,
convert = FALSE,drop = TRUE,sep = NULL){
n_val <- length(value_cols)
result <- vector(mode = "list", length = n_val)
id_cols <- setdiff(names(data), c(key_col,value_cols))
for (i in seq_along(result)){
result[[i]] <- spread(data = data[,c(id_cols,key_col,value_cols[i]),drop = FALSE],
key = !!key_col,
value = !!value_cols[i],
fill = fill,
convert = convert,
drop = drop,
sep = paste0(sep,value_cols[i],sep))
}
result %>%
purrr::reduce(.f = full_join, by = id_cols)
}
> dat %>%
spread_chr(key_col = "grp",
value_cols = c("avg","sd"),
sep = "_")
# A tibble: 2 x 5
id grp_avg_a grp_avg_b grp_sd_a grp_sd_b
<int> <dbl> <dbl> <dbl> <dbl>
1 1 1.3709584 -0.5646982 0.6569923 0.7050648
2 2 0.3631284 0.6328626 0.4577418 0.7191123
这里的关键思想是取消引用参数key_col并value_cols[i]使用!!运算符,并使用sep参数spread来控制结果值列名.
如果我们想要将此函数转换为接受键和值列的不带引号的参数,我们可以像这样修改它:
spread_nq <- function(data, key_col,..., fill = NA,
convert = FALSE, drop = TRUE, sep = NULL){
val_quos <- rlang::quos(...)
key_quo <- rlang::enquo(key_col)
value_cols <- unname(tidyselect::vars_select(names(data),!!!val_quos))
key_col <- unname(tidyselect::vars_select(names(data),!!key_quo))
n_val <- length(value_cols)
result <- vector(mode = "list",length = n_val)
id_cols <- setdiff(names(data),c(key_col,value_cols))
for (i in seq_along(result)){
result[[i]] <- spread(data = data[,c(id_cols,key_col,value_cols[i]),drop = FALSE],
key = !!key_col,
value = !!value_cols[i],
fill = fill,
convert = convert,
drop = drop,
sep = paste0(sep,value_cols[i],sep))
}
result %>%
purrr::reduce(.f = full_join,by = id_cols)
}
> dat %>%
spread_nq(key_col = grp,avg,sd,sep = "_")
# A tibble: 2 x 5
id grp_avg_a grp_avg_b grp_sd_a grp_sd_b
<int> <dbl> <dbl> <dbl> <dbl>
1 1 1.3709584 -0.5646982 0.6569923 0.7050648
2 2 0.3631284 0.6328626 0.4577418 0.7191123
这里的变化是我们使用rlang::quos和捕获不带引号的参数,rlang::enquo然后使用简单地将它们转换回字符tidyselect::vars_select.
回到使用序列的链接问题的解决方案gather,unite并且spread,我们可以使用我们学到的东西来制作这样的函数:
spread_nt <- function(data,key_col,...,fill = NA,
convert = TRUE,drop = TRUE,sep = "_"){
key_quo <- rlang::enquo(key_col)
val_quos <- rlang::quos(...)
value_cols <- unname(tidyselect::vars_select(names(data),!!!val_quos))
key_col <- unname(tidyselect::vars_select(names(data),!!key_quo))
data %>%
gather(key = ..var..,value = ..val..,!!!val_quos) %>%
unite(col = ..grp..,c(key_col,"..var.."),sep = sep) %>%
spread(key = ..grp..,value = ..val..,fill = fill,
convert = convert,drop = drop,sep = NULL)
}
> dat %>%
spread_nt(key_col = grp,avg,sd,sep = "_")
# A tibble: 2 x 5
id a_avg a_sd b_avg b_sd
* <int> <dbl> <dbl> <dbl> <dbl>
1 1 1.3709584 0.6569923 -0.5646982 0.7050648
2 2 0.3631284 0.4577418 0.6328626 0.7191123
这依赖于上一个例子中来自rlang的相同技术.我们使用一些不寻常的名称,例如..var..我们的中间变量,以减少名称与数据框中现有列冲突的可能性.
另外,我们使用sep参数unite来控制结果列名,所以在这种情况下spread我们强制使用sep = NULL.