英文:
Within dplyr::group_by, obtain the number of observations for ONE of multiple grouping variables
问题
以下是您要翻译的部分：
"It's very possible this has been asked before, however I am having a very difficult time articulating my problem.
Within my data, I have 3 variables, LOCATION, TOPIC, and RESPONSE. I would like to calculate the distribution for each combination of TOPIC and RESPONSE by LOCATION.
Create toy data and perform initial data prep
responses <- data.frame(LOCATION = c("LOC_A", "LOC_A", "LOC_A", "LOC_A", "LOC_A","LOC_A", "LOC_A", "LOC_A","LOC_B", "LOC_B", "LOC_B", "LOC_B", "LOC_B","LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C","LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C","LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C"),TOPIC = c("Dogs", "Dogs", "Dogs", "Dogs", "Dogs", "Dogs","Lizards", "Lizards", "Lizards","Lizards", "Lizards", "Lizards", "Lizards", "Lizards","Lizards", "Lizards", "Snakes", "Snakes", "Snakes", "Snakes", "Snakes","Snakes", "Dogs", "Snakes", "Dogs", "Snakes", "Dogs","Snakes", "Dogs", "Snakes"),RESP = c("Agree", "Disagree", "Agree", "Disagree", "Agree","Disagree", "Agree", "Disagree","Agree", "Disagree", "Agree", "Disagree", "Neither", "Agree","Neither", "Agree", "Neither", "Agree", "Neither","Agree", "Neither", "Agree", "Agree", "Neither","Agree", "Neither", "Agree", "Disagree", "Disagree","Neither"))
获取每个组合级别的计数
distribution <- responses %>%
table() %>%
as.data.frame() %>%
使其更易读
dplyr::arrange(LOCATION, TOPIC, RESP)
以下是一个使用循环创建所需输出的示例解决方案：
# 丑陋的循环解决方案 :(# 初始化输出容器out <- list()# 遍历每个位置for(loc in unique(distribution$LOCATION)){# 子集该位置的分布thisDist <- dplyr::filter(distribution, LOCATION == loc)# 计算该位置的每个响应的百分比thisDist$percent <- thisDist$Freq/sum(thisDist$Freq)# 存储带有百分比列的分布 dfout[[loc]] <- thisDist}# 将输出组合成单个 dfout <- do.call("rbind", out)
我想要的是一个简洁的tidyverse解决方案。以下是描述我想象中的解决方案的伪代码：
# 想象中的tidyverse解决方案 :)out <- distribution %>%group_by(LOCATION, TOPIC, RESP) %>%summarise(#percent = Freq/(sum(<all-Freq-values-for-this-group's-LOCATION-value>)))
我在这里想要做的是获取当前组的LOCATION值的所有Freq值的总和。是否有一种在group_by/summarise内部实现这一点的好方法？
感谢您的阅读，希望这不会完全令人费解。
英文:
It's very possible this has been asked before, however I am having a very difficult time articulating my problem.
Within my data, I have 3 variables, LOCATION, TOPIC, and RESPONSE. I would like to calculate the distribution for each combination of TOPIC and RESPONSE by LOCATION.
Create toy data and perform initial data prep
responses <- data.frame(LOCATION = c("LOC_A", "LOC_A", "LOC_A", "LOC_A", "LOC_A","LOC_A", "LOC_A", "LOC_A","LOC_B", "LOC_B", "LOC_B", "LOC_B", "LOC_B","LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C","LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C","LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C"),TOPIC = c("Dogs", "Dogs", "Dogs", "Dogs", "Dogs", "Dogs","Lizards", "Lizards", "Lizards","Lizards", "Lizards", "Lizards", "Lizards", "Lizards","Lizards", "Lizards", "Snakes", "Snakes", "Snakes", "Snakes", "Snakes","Snakes", "Dogs", "Snakes", "Dogs", "Snakes", "Dogs","Snakes", "Dogs", "Snakes"),RESP = c("Agree", "Disagree", "Agree", "Disagree", "Agree","Disagree", "Agree", "Disagree","Agree", "Disagree", "Agree", "Disagree", "Neither", "Agree","Neither", "Agree", "Neither", "Agree", "Neither","Agree", "Neither", "Agree", "Agree", "Neither","Agree", "Neither", "Agree", "Disagree", "Disagree","Neither"))# Obtain counts for each combination of levelsdistribution <- responses %>%table() %>%as.data.frame() %>%# Make it more readabledplyr::arrange(LOCATION, TOPIC, RESP)
Here is an example solution which uses a loop to create my desired output:
# ugly loop solution :(# Initialize output containerout <- list()# Iterate over each locationfor(loc in unique(distribution$LOCATION)){# Subset distribution for this locationthisDist <- dplyr::filter(distribution, LOCATION == loc)# Calculate percent of each response for this locationthisDist$percent <- thisDist$Freq/sum(thisDist$Freq)# Store distribution df with percent columnout[[loc]] <- thisDist}# combine output into single dfout <- do.call("rbind", out)
What I would like to have is a concise tidyverse solution. Here is some pseudo-code which describes my imaginary solution.
# Imaginary tidyverse solution :)out <- distribution %>%group_by(LOCATION, TOPIC, RESP) %>%summarise(#percent = Freq/(sum(<all-Freq-values-for-this-group's-LOCATION-value>)))
What I'm looking to do here is obtain the sum of all Freq values for the LOCATION value of the current group. Is there a nice way to do this within a group_by/summarise?
Thanks for reading, I hope this isn't completely inscrutable.
答案1
得分: 1
这是您要翻译的内容：
"Is this what you're looking for?
如果您的dplyr版本早于1.1，则使用以下代码：
distribution %>%group_by(LOCATION) %>%mutate(percent = Freq/sum(Freq))
英文:
Is this what you're looking for?
distribution %>%mutate(percent = Freq/sum(Freq), .by = LOCATION)# LOCATION TOPIC RESP Freq percent# 1 LOC_A Dogs Agree 3 0.37500000# 2 LOC_A Dogs Disagree 3 0.37500000# 3 LOC_A Dogs Neither 0 0.00000000# 4 LOC_A Lizards Agree 1 0.12500000# 5 LOC_A Lizards Disagree 1 0.12500000# 6 LOC_A Lizards Neither 0 0.00000000# 7 LOC_A Snakes Agree 0 0.00000000# 8 LOC_A Snakes Disagree 0 0.00000000# 9 LOC_A Snakes Neither 0 0.00000000# 10 LOC_B Dogs Agree 0 0.00000000# 11 LOC_B Dogs Disagree 0 0.00000000# 12 LOC_B Dogs Neither 0 0.00000000# 13 LOC_B Lizards Agree 2 0.40000000# 14 LOC_B Lizards Disagree 2 0.40000000# 15 LOC_B Lizards Neither 1 0.20000000# 16 LOC_B Snakes Agree 0 0.00000000# 17 LOC_B Snakes Disagree 0 0.00000000# 18 LOC_B Snakes Neither 0 0.00000000# 19 LOC_C Dogs Agree 3 0.17647059# 20 LOC_C Dogs Disagree 1 0.05882353# 21 LOC_C Dogs Neither 0 0.00000000# 22 LOC_C Lizards Agree 2 0.11764706# 23 LOC_C Lizards Disagree 0 0.00000000# 24 LOC_C Lizards Neither 1 0.05882353# 25 LOC_C Snakes Agree 3 0.17647059# 26 LOC_C Snakes Disagree 1 0.05882353# 27 LOC_C Snakes Neither 6 0.35294118
If you have dplyr older than 1.1, then use
distribution %>%group_by(LOCATION) %>%mutate(percent = Freq/sum(Freq))
答案2
得分: 1
The key is not to use summarise but mutate.
out <- distribution %>%
ungroup() %>%
group_by(LOCATION) %>%
mutate(percent = Freq/ sum(Freq))
英文:
The key is not to use summarise but mutate.
out <- distribution %>%ungroup() %>%group_by(LOCATION) %>%mutate(percent = Freq/ sum(Freq))
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