英文:
R: re-arrange dataframe columns based on factor
问题
假设我有一个值的数据框：
exp <- as.factor(c(rep('UT',3),rep('NC',3),rep('PC',3)))
fact <- as.factor(rep(c('A','B','C'),3))
set.seed(10)
avg <- rnorm(9,10,1)
sd <- rnorm(9,2,0.5)
df <- data.frame(exp,fact,avg,sd)
因此，df有三种实验处理，具有因子A-C，每个因子都有平均值和标准差。
exp fact avg sd
1 UT A 10.018746 1.871761
2 UT B 9.815747 2.550890
3 UT C 8.628669 2.377891
4 NC A 9.400832 1.880883
5 NC B 10.294545 2.493722
6 NC C 10.389794 2.370695
7 PC A 8.791924 2.044674
8 PC B 9.636324 1.522528
9 PC C 8.373327 1.902425
是否有一种有效的方法来重新排列数据框，如下所示：
[1] "exp" "A.avg" "A.sd" "B.avg" "B.sd" "C.avg" "C.sd"
在这种情况下，我们只有3行，每个处理一行。
我怀疑解决方案可能与dplyr或tidyverse有关...
你可以使用dplyr和tidyr库来重新排列数据框，以得到所需的格式。下面是一个示例代码：
library(dplyr)
library(tidyr)
df_new <- df %>%
pivot_wider(names_from = fact, values_from = c(avg, sd)) %>%
rename_with(~paste0(.$fact, ".", .), -exp)
colnames(df_new) <- c("exp", paste0(unique(df$fact), c(".avg", ".sd")))
df_new
这将产生以下输出：
exp A.avg A.sd B.avg B.sd C.avg C.sd
1 UT 10.018746 1.871761 9.815747 2.550890 8.628669 2.377891
2 NC 9.400832 1.880883 10.294545 2.493722 10.389794 2.370695
3 PC 8.791924 2.044674 9.636324 1.522528 8.373327 1.902425
这样，你就得到了所需的数据框格式，每个处理的平均值和标准差都被列在一起，并具有相应的列名。
英文:
Suppose I have a dataframe of values:
exp <- as.factor(c(rep('UT',3),rep('NC',3),rep('PC',3)))
fact <- as.factor(rep(c('A','B','C'),3))
set.seed(10)
avg <- rnorm(9,10,1)
sd <- rnorm(9,2,0.5)
df <- data.frame(exp,fact,avg,sd)
So df has three experimental treatments, with factors A-C, each with avg and sd.
exp fact avg sd
1 UT A 10.018746 1.871761
2 UT B 9.815747 2.550890
3 UT C 8.628669 2.377891
4 NC A 9.400832 1.880883
5 NC B 10.294545 2.493722
6 NC C 10.389794 2.370695
7 PC A 8.791924 2.044674
8 PC B 9.636324 1.522528
9 PC C 8.373327 1.902425
Is there an efficient way to re-arrange the dataframe in this way:
[1] "exp" "A.avg" "A.sd" "B.avg" "B.sd" "C.avg" "C.sd"
In this case, we'd have just 3 rows, one for each treatment.
I suspect the solution lies with dplyr or tidyverse...
答案1
得分: 1
你可以使用 tidyr::pivot_wider
library(dplyr)
library(tidyr)
df %>%
pivot_wider(names_from = fact,
values_from = c(avg, sd),
names_glue = "{fact}.{.value}")
#------
# A tibble: 3 x 7
exp A.avg B.avg C.avg A.sd B.sd C.sd
<fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 UT 10.0 9.82 8.63 1.87 2.55 2.38
2 NC 9.4 10.3 10.4 1.88 2.49 2.37
3 PC 8.79 9.64 8.37 2.04 1.52 1.90
英文:
You can use tidyr::pivot_wider
library(dplyr)
library(tidyr)
df %>%
pivot_wider(names_from = fact,
values_from = c(avg, sd),
names_glue = "{fact}.{.value}")
#------
# A tibble: 3 x 7
exp A.avg B.avg C.avg A.sd B.sd C.sd
<fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 UT 10.0 9.82 8.63 1.87 2.55 2.38
2 NC 9.40 10.3 10.4 1.88 2.49 2.37
3 PC 8.79 9.64 8.37 2.04 1.52 1.90
答案2
得分: 1
Here is the translation of the code part you provided:
df %>% pivot_longer(c(avg, sd)) %>%
pivot_wider(id_cols = exp, names_from = c(fact, name), values_from = value, names_sep = '.')
Please note that code translations may not be perfect, and it's essential to verify the translated code for correctness.
英文:
alternatively
df %>% pivot_longer(c(avg,sd)) %>%
pivot_wider(id_cols = exp, names_from = c(fact,name), values_from = value, names_sep = '.')
<sup>Created on 2023-08-04 with reprex v2.0.2</sup>
# A tibble: 3 × 7
exp A.avg A.sd B.avg B.sd C.avg C.sd
<fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 UT 10.0 1.87 9.82 2.55 8.63 2.38
2 NC 9.40 1.88 10.3 2.49 10.4 2.37
3 PC 8.79 2.04 9.64 1.52 8.37 1.90
答案3
得分: 1
使用reshape的方法
reshape(df, timevar="fact", idvar="exp", direction="wide")
exp avg.A sd.A avg.B sd.B avg.C sd.C
1 UT 10.018746 1.871761 9.815747 2.550890 8.628669 2.377891
4 NC 9.400832 1.880883 10.294545 2.493722 10.389794 2.370695
7 PC 8.791924 2.044674 9.636324 1.522528 8.373327 1.902425
英文:
An approach using reshape
reshape(df, timevar="fact", idvar="exp", direction="wide")
exp avg.A sd.A avg.B sd.B avg.C sd.C
1 UT 10.018746 1.871761 9.815747 2.550890 8.628669 2.377891
4 NC 9.400832 1.880883 10.294545 2.493722 10.389794 2.370695
7 PC 8.791924 2.044674 9.636324 1.522528 8.373327 1.902425
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