R：按此顺序对我的数据框的最小值进行排序
R: Sort the Minimums of My Data Frame in This Order
我有如下数据框：
library(future.apply)
lb <- 2:9
NBB_AR0.8 <- c(1.879, 1.065, 1.385, 1.568, 1.493, 1.732, 1.263, 0.9369)
NBB_AR0.9 <- c(0.8051, 0.7598, 1.113, 1.056, 0.9819, 0.8842, 0.679, 0.4441)
NBB_AR0.95 <- c(0.7456, 1.249, 0.8531, 1.573, 1.425, 1.181, 0.8645, 0.5171)
MBB1_AR0.8 <- c(1.806, 1.611, 1.199, 1.46, 1.253, 1.483, 1.418,1.615)
MBB1_AR0.9 <- c(0.7936, 0.7351, 0.9151, 0.9417, 0.9827, 0.9767, 0.8699, 0.9629)
MBB1_AR0.95 <- c(1.646, 1.621, 0.9941, 0.9044, 1.054, 1.247, 1.376, 1.281)
MBB2_AR0.8 <- c(1.806, 1.611, 1.199, 1.46, 1.619, 1.483, 1.498, 1.301)
MBB2_AR0.9 <- c(0.7936, 0.7351, 0.9151, 0.9417, 0.9653, 0.9767, 1.051, 0.9979)
MBB2_AR0.95 <- c(1.646, 1.621, 0.9941, 0.9044, 1.531, 1.247, 1.03, 0.9696)
MBB3_AR0.8 <- c(1.806, 1.611, 1.199, 1.46, 1.363, 1.483, 1.742, 1.161)
MBB3_AR0.9 <- c(0.7936, 0.7351, 0.9151, 0.9417, 1.025, 0.9767, 0.9018, 0.6612)
MBB3_AR0.95 <- c(1.646, 1.621, 0.9941, 0.9044, 0.861, 1.247, 1.184, 0.8825)
CBB_AR0.8 <- c(1.642, 0.9616, 1.42, 1.728, 1.326, 1.324, 1.542, 1.172)
CBB_AR0.9 <- c(0.2077, 0.2158, 0.1791, 0.1933, 0.168, 0.2211, 0.1516, 0.2133)
CBB_AR0.95 <- c(0.1039, 0.08983, 0.09176, 0.1, 0.09203, 0.08383, 0.08386, 0.08956)
df <- data.frame(lb, NBB_AR0.8, NBB_AR0.9, NBB_AR0.95, NBB_AR0.95, MBB1_AR0.8, MBB1_AR0.9, MBB1_AR0.95, MBB2_AR0.8, MBB2_AR0.9, MBB2_AR0.95, MBB3_AR0.8, MBB3_AR0.9, MBB3_AR0.95, CBB_AR0.8, CBB_AR0.9, CBB_AR0.95)
- 向量NBB_AR0.8的最小值是
min(NBB_AR0.8) = 0.9369
- 向量NBB_AR0.9的最小值为
min(NBB_AR0.9) = 0.4441
- 向量NBB_AR0.95的最小值为
min(NBB_AR0.95) = 0.5171
以上三（3）个共有NBB，因此应排在NBB
行
- 向量NBB_AR0.8的最小值是
min(NBB_AR0.8) = 0.9369
- 向量MBB1_AR0.8的最小值是
min(MBB1_AR0.8) = 1.199
- 向量MBB2_AR0.8的最小值是
min(MBB2_AR0.8) = 1.199
- 向量MBB3_AR0.8的最小值为
min(MBB3_AR0.8) = 1.161
- 向量CBB_AR0.8的最小值为
min(CBB_AR0.8) = 0.9616
以上五（5）个共有AR0.8个，因此应排在AR0.8行
其他的也是一样的安排。
我希望使用 R 将最小值排列如下：
AR0.8
AR0.9
AR0.95
NBB
0.9369
0.4441
0.5171
MBB1
1.199
0.7351
0.9044
MBB2
1.199
0.7351
0.9044
MBB3
1.161
0.6612
0.861
CBB
0.9616
0.1516
0.08336
我试过了，但得到的结果不符合我对安排的期望：
future.apply::future_apply(df[-1], 2, min)
> NBB_AR0.8 NBB_AR0.9 NBB_AR0.95 NBB_AR0.95.1 MBB1_AR0.8 MBB1_AR0.9 MBB1_AR0.95 MBB2_AR0.8 MBB2_AR0.9 MBB2_AR0.95 MBB3_AR0.8
0.93690 0.44410 0.51710 0.51710 1.19900 0.73510 0.90440 1.19900 0.73510 0.90440 1.16100
MBB3_AR0.9 MBB3_AR0.95 CBB_AR0.8 CBB_AR0.9 CBB_AR0.95
0.66120 0.86100 0.96160 0.15160 0.08383
回答正确，但我也对编排感兴趣
我也对这个方法感兴趣：
future.apply::future_apply(df[-1], 2, which.min)
这给了我这个：
NBB_N10_AR0.8_RMSE NBB_N10_AR0.9_RMSE NBB_N10_AR0.95_RMSE NBB_N10_AR0.95_RMSE.1 MBB1_N10_AR0.8_RMSE MBB1_N10_AR0.9_RMSE
8 8 8 8 3 2
MBB1_N10_AR0.95_RMSE MBB2_N10_AR0.8_RMSE MBB2_N10_AR0.9_RMSE MBB2_N10_AR0.95_RMSE MBB3_N10_AR0.8_RMSE MBB3_N10_AR0.9_RMSE
4 3 2 4 8 8
MBB3_N10_AR0.95_RMSE CBB_N10_AR0.8_RMSE CBB_N10_AR0.9_RMSE CBB_N10_AR0.95_RMSE
5 2 7 6
我要这样排列table:
AR0.8
AR0.9
AR0.95
NBB
9
9
9
MBB1
4
3
5
MBB2
4
3
5
MBB3
9
9
6
CBB
3
8
8
- 向量NBB_AR0.8的最小值是
lb = 9 下的min(NBB_AR0.8) = 0.9369
- 向量NBB_AR0.9的最小值是
lb = 9 下的min(NBB_AR0.9) = 0.4441
- 向量NBB_AR0.95的最小值是
lb = 9 下的min(NBB_AR0.95) = 0.5171
以上三（3）个共有NBB，因此应排在NBB
行
- 向量NBB_AR0.8的最小值是
lb = 9 下的min(NBB_AR0.8) = 0.9369
- 向量MBB1_AR0.8的最小值是
lb = 4 下的min(MBB1_AR0.8) = 1.199
- 向量MBB2_AR0.8的最小值是
lb = 4 下的min(MBB2_AR0.8) = 1.199
- 向量MBB3_AR0.8的最小值是
min(MBB3_AR0.8) = 1.161下的lb = 9
- 向量CBB_AR0.8的最小值是
lb = 3下的min(CBB_AR0.8) = 0.9616
以上五（5）个共有AR0.8个，因此应排在AR0.8
行
我们可以使用
lst1 <- split(setNames(out, sub(".*_", "", names(out))), sub("_.*", "", names(out)))
do.call(rbind, lapply(lst1, function(x) x[!duplicated(x)]))
-输出
AR0.8 AR0.9 AR0.95
CBB 0.9616 0.1516 0.08383
MBB1 1.1990 0.7351 0.90440
MBB2 1.1990 0.7351 0.90440
MBB3 1.1610 0.6612 0.86100
NBB 0.9369 0.4441 0.51710
lst2 <- split(setNames(out2, sub(".*_", "", names(out2))), sub("_.*", "", names(out2)))
do.call(rbind, lapply(lst2, `[`, 1:3))
AR0.8 AR0.9 AR0.95
CBB 2 7 6
MBB1 3 2 4
MBB2 3 2 4
MBB3 8 8 5
NBB 8 8 8
数据
out <- future.apply::future_apply(df[-1], 2, min)
out2 <- future.apply::future_apply(df[-1], 2, which.min)
一个tidyverse解决方案可以是
library(tidyr)
library(dplyr)
df %>%
pivot_longer(-c(lb), names_to = c("name", "name2"), names_pattern = "(.*)_(.*)") %>%
select(-lb) %>%
group_by(name, name2) %>%
slice_min(value) %>%
pivot_wider(names_from = name2) %>%
ungroup()
回归
# A tibble: 5 x 4
name AR0.8 AR0.9 AR0.95
<chr> <dbl> <dbl> <dbl>
1 CBB 0.962 0.152 0.0838
2 MBB1 1.20 0.735 0.904
3 MBB2 1.20 0.735 0.904
4 MBB3 1.16 0.661 0.861
5 NBB 0.937 0.444 0.517
我有如下数据框：
library(future.apply)
lb <- 2:9
NBB_AR0.8 <- c(1.879, 1.065, 1.385, 1.568, 1.493, 1.732, 1.263, 0.9369)
NBB_AR0.9 <- c(0.8051, 0.7598, 1.113, 1.056, 0.9819, 0.8842, 0.679, 0.4441)
NBB_AR0.95 <- c(0.7456, 1.249, 0.8531, 1.573, 1.425, 1.181, 0.8645, 0.5171)
MBB1_AR0.8 <- c(1.806, 1.611, 1.199, 1.46, 1.253, 1.483, 1.418,1.615)
MBB1_AR0.9 <- c(0.7936, 0.7351, 0.9151, 0.9417, 0.9827, 0.9767, 0.8699, 0.9629)
MBB1_AR0.95 <- c(1.646, 1.621, 0.9941, 0.9044, 1.054, 1.247, 1.376, 1.281)
MBB2_AR0.8 <- c(1.806, 1.611, 1.199, 1.46, 1.619, 1.483, 1.498, 1.301)
MBB2_AR0.9 <- c(0.7936, 0.7351, 0.9151, 0.9417, 0.9653, 0.9767, 1.051, 0.9979)
MBB2_AR0.95 <- c(1.646, 1.621, 0.9941, 0.9044, 1.531, 1.247, 1.03, 0.9696)
MBB3_AR0.8 <- c(1.806, 1.611, 1.199, 1.46, 1.363, 1.483, 1.742, 1.161)
MBB3_AR0.9 <- c(0.7936, 0.7351, 0.9151, 0.9417, 1.025, 0.9767, 0.9018, 0.6612)
MBB3_AR0.95 <- c(1.646, 1.621, 0.9941, 0.9044, 0.861, 1.247, 1.184, 0.8825)
CBB_AR0.8 <- c(1.642, 0.9616, 1.42, 1.728, 1.326, 1.324, 1.542, 1.172)
CBB_AR0.9 <- c(0.2077, 0.2158, 0.1791, 0.1933, 0.168, 0.2211, 0.1516, 0.2133)
CBB_AR0.95 <- c(0.1039, 0.08983, 0.09176, 0.1, 0.09203, 0.08383, 0.08386, 0.08956)
df <- data.frame(lb, NBB_AR0.8, NBB_AR0.9, NBB_AR0.95, NBB_AR0.95, MBB1_AR0.8, MBB1_AR0.9, MBB1_AR0.95, MBB2_AR0.8, MBB2_AR0.9, MBB2_AR0.95, MBB3_AR0.8, MBB3_AR0.9, MBB3_AR0.95, CBB_AR0.8, CBB_AR0.9, CBB_AR0.95)
- 向量NBB_AR0.8的最小值是
min(NBB_AR0.8) = 0.9369 - 向量NBB_AR0.9的最小值为
min(NBB_AR0.9) = 0.4441 - 向量NBB_AR0.95的最小值为
min(NBB_AR0.95) = 0.5171
以上三（3）个共有NBB，因此应排在NBB
- 向量NBB_AR0.8的最小值是
min(NBB_AR0.8) = 0.9369 - 向量MBB1_AR0.8的最小值是
min(MBB1_AR0.8) = 1.199 - 向量MBB2_AR0.8的最小值是
min(MBB2_AR0.8) = 1.199 - 向量MBB3_AR0.8的最小值为
min(MBB3_AR0.8) = 1.161 - 向量CBB_AR0.8的最小值为
min(CBB_AR0.8) = 0.9616
以上五（5）个共有AR0.8个，因此应排在AR0.8行
其他的也是一样的安排。
我希望使用 R 将最小值排列如下：
| AR0.8 | AR0.9 | AR0.95 | |
|---|---|---|---|
| NBB | 0.9369 | 0.4441 | 0.5171 |
| MBB1 | 1.199 | 0.7351 | 0.9044 |
| MBB2 | 1.199 | 0.7351 | 0.9044 |
| MBB3 | 1.161 | 0.6612 | 0.861 |
| CBB | 0.9616 | 0.1516 | 0.08336 |
我试过了，但得到的结果不符合我对安排的期望：
future.apply::future_apply(df[-1], 2, min)
> NBB_AR0.8 NBB_AR0.9 NBB_AR0.95 NBB_AR0.95.1 MBB1_AR0.8 MBB1_AR0.9 MBB1_AR0.95 MBB2_AR0.8 MBB2_AR0.9 MBB2_AR0.95 MBB3_AR0.8
0.93690 0.44410 0.51710 0.51710 1.19900 0.73510 0.90440 1.19900 0.73510 0.90440 1.16100
MBB3_AR0.9 MBB3_AR0.95 CBB_AR0.8 CBB_AR0.9 CBB_AR0.95
0.66120 0.86100 0.96160 0.15160 0.08383
回答正确，但我也对编排感兴趣
我也对这个方法感兴趣：
future.apply::future_apply(df[-1], 2, which.min)
这给了我这个：
NBB_N10_AR0.8_RMSE NBB_N10_AR0.9_RMSE NBB_N10_AR0.95_RMSE NBB_N10_AR0.95_RMSE.1 MBB1_N10_AR0.8_RMSE MBB1_N10_AR0.9_RMSE 8 8 8 8 3 2 MBB1_N10_AR0.95_RMSE MBB2_N10_AR0.8_RMSE MBB2_N10_AR0.9_RMSE MBB2_N10_AR0.95_RMSE MBB3_N10_AR0.8_RMSE MBB3_N10_AR0.9_RMSE 4 3 2 4 8 8 MBB3_N10_AR0.95_RMSE CBB_N10_AR0.8_RMSE CBB_N10_AR0.9_RMSE CBB_N10_AR0.95_RMSE 5 2 7 6
我要这样排列table:
| AR0.8 | AR0.9 | AR0.95 | |
|---|---|---|---|
| NBB | 9 | 9 | 9 |
| MBB1 | 4 | 3 | 5 |
| MBB2 | 4 | 3 | 5 |
| MBB3 | 9 | 9 | 6 |
| CBB | 3 | 8 | 8 |
- 向量NBB_AR0.8的最小值是
lb = 9下的 - 向量NBB_AR0.9的最小值是
lb = 9下的 - 向量NBB_AR0.95的最小值是
lb = 9下的
min(NBB_AR0.8) = 0.9369
min(NBB_AR0.9) = 0.4441
min(NBB_AR0.95) = 0.5171
以上三（3）个共有NBB，因此应排在NBB
- 向量NBB_AR0.8的最小值是
lb = 9下的 - 向量MBB1_AR0.8的最小值是
lb = 4下的 - 向量MBB2_AR0.8的最小值是
lb = 4下的 - 向量MBB3_AR0.8的最小值是
min(MBB3_AR0.8) = 1.161下的lb = 9 - 向量CBB_AR0.8的最小值是
lb = 3下的min(CBB_AR0.8) = 0.9616
min(NBB_AR0.8) = 0.9369
min(MBB1_AR0.8) = 1.199
min(MBB2_AR0.8) = 1.199
以上五（5）个共有AR0.8个，因此应排在AR0.8
我们可以使用
lst1 <- split(setNames(out, sub(".*_", "", names(out))), sub("_.*", "", names(out)))
do.call(rbind, lapply(lst1, function(x) x[!duplicated(x)]))
-输出
AR0.8 AR0.9 AR0.95
CBB 0.9616 0.1516 0.08383
MBB1 1.1990 0.7351 0.90440
MBB2 1.1990 0.7351 0.90440
MBB3 1.1610 0.6612 0.86100
NBB 0.9369 0.4441 0.51710
lst2 <- split(setNames(out2, sub(".*_", "", names(out2))), sub("_.*", "", names(out2)))
do.call(rbind, lapply(lst2, `[`, 1:3))
AR0.8 AR0.9 AR0.95
CBB 2 7 6
MBB1 3 2 4
MBB2 3 2 4
MBB3 8 8 5
NBB 8 8 8
数据
out <- future.apply::future_apply(df[-1], 2, min)
out2 <- future.apply::future_apply(df[-1], 2, which.min)
一个tidyverse解决方案可以是
library(tidyr)
library(dplyr)
df %>%
pivot_longer(-c(lb), names_to = c("name", "name2"), names_pattern = "(.*)_(.*)") %>%
select(-lb) %>%
group_by(name, name2) %>%
slice_min(value) %>%
pivot_wider(names_from = name2) %>%
ungroup()
回归
# A tibble: 5 x 4
name AR0.8 AR0.9 AR0.95
<chr> <dbl> <dbl> <dbl>
1 CBB 0.962 0.152 0.0838
2 MBB1 1.20 0.735 0.904
3 MBB2 1.20 0.735 0.904
4 MBB3 1.16 0.661 0.861
5 NBB 0.937 0.444 0.517