对于数据帧，我需要按行查找从第2列开始的未知列数的最小值和最大值。这是一个示例：
library(tidyverse)
# test data
(test_data <- tibble(id = c(1:9),
x = runif(9),
x2 = runif(9),
x3 = runif(9)))
samples = 100
# This example, which specifies the column names, correctly finds the min and max values by row
(test_1 <- test_data %>%
rowwise() %>%
mutate(min_val = min(x, x2, x3), max_val = max(x, x2, x3)))
# This example does not
(test_2 <- test_data %>%
rowwise() %>%
mutate(min_val = min(x:x3), max_val = max(x:x3)))
我实际上想做的是
mutate(min_val = min([,2:samples+1]), max_val = max([,2:samples+1])))
因为（1）我希望保留id列（以供以后与另一个数据框连接），并且（2）按列位置指定似乎是实现此目的的一种明显方法，因为我对列名和样本都不关心大。
谢谢！
编辑后的示例
这（建议）
test_data %>%
nest(-id) %>% # nest rest of columns apart from id
mutate(min_val = map(data, min), # get min and max
max_val = map(data, max)) %>%
unnest()
处理原始测试数据。但是，现实世界中的数据具有重复的id，例如
(test_data <- tibble(id = c(1:9, 1:9),
x = runif(18),
x2 = runif(18),
x3 = runif(18)))
并导致“错误：所有嵌套列必须具有相同数量的元素。”
答案 0 :(得分：1)
一种可能的tidyverse解决方案是nest除id以外的任何列，然后使用map获得min和{{1 }}。您无需指定任何列名：
max在具有多个ID的情况下，您可以使用以下方法：
library(tidyverse)
# test data
(test_data <- tibble(id = c(1:9),
x = runif(9),
x2 = runif(9),
x3 = runif(9)))
samples = 100
test_data %>%
nest(-id) %>% # nest rest of columns apart from id
mutate(min_val = map(data, min), # get min and max
max_val = map(data, max)) %>%
unnest() # unnest columns
# # A tibble: 9 x 6
# id min_val max_val x x2 x3
# <int> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 1 0.0217 0.239 0.130 0.0217 0.239
# 2 2 0.125 0.814 0.625 0.814 0.125
# 3 3 0.281 0.770 0.331 0.770 0.281
# 4 4 0.123 0.868 0.123 0.644 0.868
# 5 5 0.149 0.340 0.149 0.340 0.337
# 6 6 0.496 0.865 0.596 0.865 0.496
# 7 7 0.0766 0.984 0.0766 0.656 0.984
# 8 8 0.272 0.926 0.702 0.926 0.272
# 9 9 0.433 0.912 0.912 0.433 0.590
答案 1 :(得分：0)
如果要根据列名选择列：
library(tidyverse)
df <- tibble(id = c(1:9),
x = runif(9),
x2 = runif(9),
x3 = runif(9))
df <- tibble::rowid_to_column(df, "id_uni") #Assigning a unique ID to all rows
df2 <- df %>%
group_by(id_uni) %>%
do(min_val = min(select(., grep("x", names(df), value = TRUE))),
max_val = max(select(., grep("x", names(df), value = TRUE)))) %>%
unnest()
df <- left_join(df, df2, by = "id_uni")
# A tibble: 9 x 7
id_uni id x x2 x3 min_val max_val
<int> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 0.714 0.879 0.943 0.714 0.943
2 2 2 0.240 0.409 0.375 0.240 0.409
3 3 3 0.896 0.205 0.00804 0.00804 0.896
4 4 4 0.483 0.471 0.981 0.471 0.981
5 5 5 0.263 0.379 0.378 0.263 0.379
6 6 6 0.320 0.248 0.986 0.248 0.986
7 7 7 0.664 0.925 0.140 0.140 0.925
8 8 8 0.728 0.466 0.562 0.466 0.728
9 9 9 0.571 0.0139 0.794 0.0139 0.794
或者：
df2 <- df %>%
group_by(id_uni) %>%
do(min_val = min(select(.,starts_with("x"))),
max_val = max(select(.,starts_with("x")))) %>%
unnest()
df <- left_join(df, df2, by = "id_uni")
# A tibble: 9 x 7
id_uni id x x2 x3 min_val max_val
<int> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 0.101 0.606 0.743 0.101 0.743
2 2 2 0.312 0.496 0.0684 0.0684 0.496
3 3 3 0.434 0.109 0.532 0.109 0.532
4 4 4 0.616 0.846 0.123 0.123 0.846
5 5 5 0.339 0.446 0.676 0.339 0.676
6 6 6 0.580 0.750 0.0560 0.0560 0.750
7 7 7 0.830 0.796 0.798 0.796 0.830
8 8 8 0.335 0.391 0.663 0.335 0.663
9 9 9 0.382 0.0148 0.244 0.0148 0.382
答案 2 :(得分：0)
这是pmin/pmax
library(tidyverse)
test_data %>%
mutate(min_val = pmin(!!! rlang::syms(names(.)[-1])),
max_val = pmax(!!! rlang::syms(names(.)[-1])))
# A tibble: 9 x 6
# id x x2 x3 min_val max_val
# <int> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 0.293 0.255 0.501 0.255 0.501
#2 2 0.225 0.605 0.139 0.139 0.605
#3 3 0.704 0.371 0.0939 0.0939 0.704
#4 4 0.519 0.672 0.552 0.519 0.672
#5 5 0.663 0.673 0.725 0.663 0.725
#6 6 0.920 0.320 0.138 0.138 0.920
#7 7 0.280 0.904 0.223 0.223 0.904
#8 8 0.764 0.198 0.688 0.198 0.764
#9 9 0.802 0.0442 0.0765 0.0442 0.802
set.seed(24)
test_data <- tibble(id = c(1:9),
x = runif(9),
x2 = runif(9),
x3 = runif(9))