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 sd1 UT A 10.018746 1.8717612 UT B 9.815747 2.5508903 UT C 8.628669 2.3778914 NC A 9.400832 1.8808835 NC B 10.294545 2.4937226 NC C 10.389794 2.3706957 PC A 8.791924 2.0446748 PC B 9.636324 1.5225289 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.sd1 UT 10.018746 1.871761 9.815747 2.550890 8.628669 2.3778912 NC 9.400832 1.880883 10.294545 2.493722 10.389794 2.3706953 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 sd1 UT A 10.018746 1.8717612 UT B 9.815747 2.5508903 UT C 8.628669 2.3778914 NC A 9.400832 1.8808835 NC B 10.294545 2.4937226 NC C 10.389794 2.3706957 PC A 8.791924 2.0446748 PC B 9.636324 1.5225289 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 7exp 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 2.382 NC 9.4 10.3 10.4 1.88 2.49 2.373 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 7exp 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 2.382 NC 9.40 10.3 10.4 1.88 2.49 2.373 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 × 7exp 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 2.382 NC 9.40 1.88 10.3 2.49 10.4 2.373 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.C1 UT 10.018746 1.871761 9.815747 2.550890 8.628669 2.3778914 NC 9.400832 1.880883 10.294545 2.493722 10.389794 2.3706957 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.C1 UT 10.018746 1.871761 9.815747 2.550890 8.628669 2.3778914 NC 9.400832 1.880883 10.294545 2.493722 10.389794 2.3706957 PC 8.791924 2.044674 9.636324 1.522528 8.373327 1.902425
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论