作者ddtwu (<囧>真夭寿)
看板R_Language
标题Re: [问题] 如何将一个栏位按降幂排列变成另一个栏位
时间Sun Aug 13 21:24:06 2017
※ 引述《celestialgod (天)》之铭言：
: ※ 引述《henry48124 (= =)》之铭言：
: : [问题类型]:
: : 程式谘询(我想用R 做某件事情，但是我不知道要怎麽用R 写出来)
: : [软体熟悉度]:
: : 入门(写过其他程式，只是对语法不熟悉)
: : [问题叙述]:
: : 各位大大好，我有一笔资料长得像是：
: : head(df)
: : id place count
: : 1 A 1
: : 1 B 1
: : 2 B 1
: : 2 C 3
: : 3 D 2
: : 4 A 1
: : 4 C 2
: : 4 D 5
: : 5 B 1
: : 我希望能让他变成
: : id count top_place1 top_place2
: : 1 2 A B
: : 2 4 C B
: : 3 2 D
: : 4 8 D C
: : 5 1 B
: : [程式范例]:
: : 这是我目前的做法，总觉得写得怪怪的，如果未来要做到 top100 就不能这样写
: : 谢谢各位 Orz
: : library(dplyr)
: : answer <- NULL
: : for(x in as.list(unique(df$id))) {
: : df_id <- df %>%
: : filter(id == x) %>%
: : arrange(-count)
: : count <- sum(df$count)
: : top_place1 <- NA
: : top_place2 <- NA
: : col <- c(x, count, top_place1, top_place2)
: : for(y in 1:nrow(df_id)) {
: : if(y <= 2) {
: : col[y+2] <- df_id[y,]$place
: : }
: : answer <- rbind(answer, col)
: : }
: : [环境叙述]:
: : [关键字]:
: method 1是硬干，可以直接先看method 2
: library(data.table)
: library(stringr)
: library(pipeR)
: DT <- data.table(id = rep(1:5, c(2,2,1,3,1)),
: place = c("A","B","B","C","D","A","C","D","B"),
: count = c(1,1,1,3:1,2,5,1))
: ## method 1:
: setorder(DT, id, -count, -place)
: numRank <- 3
: DT[ , .(lapply(1:numRank, function(i){
: ifelse(length(place) >= i, place[i], "")
: }) %>>% transpose %>>% sapply(str_c, collapse = ",")), by = .(id)] %>>%
: `[`(j = str_c("top_place", 1:numRank) := transpose(str_split(V1, ",")),
: by = .(id)) %>>%
: `[`(j = V1 := NULL) %>>%
: merge(DT[ , .(count = sum(count)), by = .(id)], by = "id")
: # id top_place1 top_place2 top_place3 count
: # 1: 1 A B 2
: # 2: 2 C B 4
: # 3: 3 D 2
: # 4: 4 D C A 8
: # 5: 5 B 1
: ## method 2:
: setorder(DT, id, -count, -place)
: numRank <- 3
: DT[ , rr := length(count) - frank(count, ties.method = "first")+1, by = .(id)]
: DT[rr %in% 1:numRank] %>>%
: dcast(id ~ rr, value.var = "place") %>>%
: setnames(as.character(1:numRank), str_c("top_place", 1:numRank)) %>>%
: merge(DT[ , .(count = sum(count)), by = .(id)], by = "id")
: # id top_place1 top_place2 top_place3 count
: # 1: 1 A B NA 2
: # 2: 2 C B NA 4
: # 3: 3 D NA NA 2
: # 4: 4 D C A 8
: # 5: 5 B NA NA 1
我的作法是这样：
library(dplyr)
library(magrittr)
library(tidyr)
df <- data.frame(id = rep(1:5, c(2,2,1,3,1)),
place = c("A","B","B","C","D","A","C","D","B"),
count = c(1,1,1,3:1,2,5,1), stringsAsFactors = FALSE) %>% tbl_df()
df %>% group_by(id) %>%
mutate(seq = order(count, decreasing = TRUE),
sumCount = sum(count)) %>%
filter(seq <= 2) %>%
ungroup() %>%
mutate(seqName = sprintf('top_place%s', seq)) %>%
select(-count, -seq) %>%
spread(key = seqName, value = place, fill = NA)
# A tibble: 5 x 4
id sumCount top_place1 top_place2
* <int> <dbl> <chr> <chr>
1 1 2 A B
2 2 4 C B
3 3 2 D <NA>
4 4 8 D C
5 5 1 B <NA>
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