英文:
R: Merge information from 2 columns together
问题
我已创建了一个示例数据框，其中包含3个不同的组，每个组有2列。
Group_1显示参与者的总数，Group_1_Pos显示总参与者中有多少人是积极的，依此类推：
df1 <- structure(list(Date = c("2016", "2017", "2018", "2019"),Group_1 = c("100", "200", "300", "400"),Group_1_Pos = c("10", "20", "30", "40"),Group_2 = c("500", "600", "700", "800"),Group_2_Pos = c("50", "60", "70", "80"),Group_3 = c("900", "1000", "1100", "1200"),Group_3_Pos = c("90", "100", "110", "120")),class = "data.frame", row.names=c("1", "2", "3", "4"))> df1Date Group_1 Group_1_Pos Group_2 Group_2_Pos Group_3 Group_3_Pos1 2016 100 10 500 50 900 902 2017 200 20 600 60 1000 1003 2018 300 30 700 70 1100 1104 2019 400 40 800 80 1200 120
我想将总参与者列与积极参与者列合并，以保持两个值仍然用括号分开。例如：
Date Group_1 Group_2 Group_31 2016 100 (10) 500 (50) 900 (90)2 2017 200 (20) 600 (60) 1000 (100)3 2018 300 (30) 700 (70) 1100 (110)4 2019 400 (40) 800 (80) 1200 (120)
因此，在这个示例中，我将积极的参与者添加到总参与者旁边，并仅保留3列用于3个组。
英文:
I have created an example dataframe which has 3 different groups with 2 columns for each group.
Group_1 shows the total amount of participants and Group_1_Pos shows how many of the total participants are positive, etc:
df1 <- structure(list(Date = c("2016", "2017", "2018", "2019"),Group_1 = c("100", "200", "300", "400"),Group_1_Pos = c("10", "20", "30", "40"),Group_2 = c("500", "600", "700", "800"),Group_2_Pos = c("50", "60", "70", "80"),Group_3 = c("900", "1000", "1100", "1200"),Group_3_Pos = c("90", "100", "110", "120")),class = "data.frame", row.names=c("1", "2", "3", "4"))> df1Date Group_1 Group_1_Pos Group_2 Group_2_Pos Group_3 Group_3_Pos1 2016 100 10 500 50 900 902 2017 200 20 600 60 1000 1003 2018 300 30 700 70 1100 1104 2019 400 40 800 80 1200 120
I would like to combine the total participant columns together with the positive participant columns in a way that keeps both values still seperated with brackets. As an example:
Date Group_1 Group_2 Group_31 2016 100 (10) 500 (50) 900 (90)2 2017 200 (20) 600 (60) 1000 (100)3 2018 300 (30) 700 (70) 1100 (110)4 2019 400 (40) 800 (80) 1200 (120)
So in this example I add the positive participants in () brackets next to the total participants and only keep 3 columns for the 3 groups.
Any help would be appreciated.
答案1
得分: 1
使用 dplyr，您可以尝试以下方式：
library(dplyr)df1 %>%mutate(Group_1 = paste0(Group_1, " (", Group_1_Pos, ")"),Group_2 = paste0(Group_2, " (", Group_2_Pos, ")"),Group_3 = paste0(Group_3, " (", Group_3_Pos, ")"),) %>%select(-contains("Pos"))# Date Group_1 Group_2 Group_3# 1 2016 100 (10) 500 (50) 900 (90)# 2 2017 200 (20) 600 (60) 1000 (100)# 3 2018 300 (30) 700 (70) 1100 (110)# 4 2019 400 (40) 800 (80) 1200 (120)
英文:
Using dplyr you could go for something like:
library(dplyr)df1 %>%mutate(Group_1 = paste0(Group_1, " (", Group_1_Pos, ")"),Group_2 = paste0(Group_2, " (", Group_2_Pos, ")"),Group_3 = paste0(Group_3, " (", Group_3_Pos, ")"),) %>%select(-contains("Pos"))# Date Group_1 Group_2 Group_3# 1 2016 100 (10) 500 (50) 900 (90)# 2 2017 200 (20) 600 (60) 1000 (100)# 3 2018 300 (30) 700 (70) 1100 (110)# 4 2019 400 (40) 800 (80) 1200 (120)
答案2
得分: 1
A purrr-dplyr-stringr:
other_values <- df1[, seq(1, ncol(df1), 2)]df1 %>%select(-contains("Pos")) %>%purrr::map2_df(., other_values, function(x, y) paste0(x, " (", y, ")")) %>%mutate(Date = stringr::str_remove_all(Date, "\\s.*"))
A tibble: 4 x 4
Date Group_1 Group_2 Group_3
1 2016 100 (10) 500 (50) 900 (90)
2 2017 200 (20) 600 (60) 1000 (100)
3 2018 300 (30) 700 (70) 1100 (110)
4 2019 400 (40) 800 (80) 1200 (120)
<details><summary>英文:</summary>A `purrr`-`dplyr`-`stringr`:other_values <- df1[,seq(1,ncol(df1),2)]df1 %>%select(-contains("Pos")) %>%purrr::map2_df(.,other_values,function(x,y) paste0(x," (",y,")")) %>%mutate(Date=stringr::str_remove_all(Date,"\\s.*"))# A tibble: 4 x 4Date Group_1 Group_2 Group_3<chr> <chr> <chr> <chr>1 2016 100 (10) 500 (50) 900 (90)2 2017 200 (20) 600 (60) 1000 (100)3 2018 300 (30) 700 (70) 1100 (110)4 2019 400 (40) 800 (80) 1200 (120)</details># 答案3**得分**: 1以下是翻译好的代码部分：```R这是一个使用基本R方式来完成问题要求的方法。使用正则表达式和`grep`获取要粘贴的列，然后遍历索引向量并将它们粘贴在一起。最后，使用`cbind`将第一列和这个结果合并。inx <- grep("\\d$", names(df1))tmp <- sapply(inx, function(i) paste(df1[[i]], paste0("(", df1[[i + 1]], ")")))res <- cbind(df1[1], tmp)names(res)[-1] <- names(df1)[inx]res# Date Group_1 Group_2 Group_3#1 2016 100 (10) 500 (50) 900 (90)#2 2017 200 (20) 600 (60) 1000 (100)#3 2018 300 (30) 700 (70) 1100 (110)#4 2019 400 (40) 800 (80) 1200 (120)最后清理。rm(inx, tmp)
英文:
Here is a base R way of doing what the question asks for.
Get the columns to be pasted with a regex and grep, then loop through the indices vector and paste them together. Finally, cbind the first column and this result.
inx <- grep("\\d$", names(df1))tmp <- sapply(inx, function(i) paste(df1[[i]], paste0("(", df1[[i + 1]], ")")))res <- cbind(df1[1], tmp)names(res)[-1] <- names(df1)[inx]res# Date Group_1 Group_2 Group_3#1 2016 100 (10) 500 (50) 900 (90)#2 2017 200 (20) 600 (60) 1000 (100)#3 2018 300 (30) 700 (70) 1100 (110)#4 2019 400 (40) 800 (80) 1200 (120)
Final clean up.
rm(inx, tmp)
答案4
得分: 1
给定3个组，这是一个基于R语言的解决方案，可以为您提供所需的输出：
n <- 3dfout <- cbind(df1[1],`colnames<-`(sapply(seq(n), function(k) paste0(df[[x <- paste0("Group_",k)]], " (", df[[paste0(x,"_Pos")]], ")")),paste0("Group", seq(n))))
结果如下：
> dfoutDate Group1 Group2 Group31 2016 100 (10) 500 (50) 900 (90)2 2017 200 (20) 600 (60) 1000 (100)3 2018 300 (30) 700 (70) 1100 (110)4 2019 400 (40) 800 (80) 1200 (120)
如果您有任何其他问题，请随时提出。
英文:
Given 3 groups, here is a base R solution that can give you the desired output
n <- 3dfout <- cbind(df1[1],`colnames<-`(sapply(seq(n), function(k) paste0(df[[x <- paste0("Group_",k)]]," (", df[[paste0(x,"_Pos")]],")")),paste0("Group",seq(n))))
such that
> dfoutDate Group1 Group2 Group31 2016 100 (10) 500 (50) 900 (90)2 2017 200 (20) 600 (60) 1000 (100)3 2018 300 (30) 700 (70) 1100 (110)4 2019 400 (40) 800 (80) 1200 (120)
答案5
得分: 0
这是一个更通用的tidyverse解决方案
library(tidyverse)df1 %>%rename_at(vars(contains("Pos")),~ str_remove(., "_Pos") %>%str_remove("Group_") %>%str_c("Pos", ., sep = "_")) %>%pivot_longer(Group_1:Pos_3,names_to = c(".value", "set"),names_sep = "_") %>%mutate(Pos = Pos %>%str_c("(", ., ")")) %>%unite("result", Group:Pos, sep = "") %>%pivot_wider(names_from = set, values_from = result)
请注意，这是R代码的翻译。
英文:
Here is a more general tidyverse solution
library(tidyverse)df1 %>%rename_at(vars(contains("Pos")),~ str_remove(., "_Pos") %>% str_remove("Group_") %>% str_c("Pos", ., sep = "_")) %>%pivot_longer(Group_1:Pos_3,names_to = c(".value", "set"),names_sep = "_") %>%mutate(Pos = Pos %>% str_c("(", ., ")")) %>%unite("result", Group:Pos, sep = "") %>%pivot_wider(names_from = set, values_from = result)
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