我遇到了一个困难的数据帧问题，我试图从现有的数据帧中创建新的列/列名称/列值，这些数据帧的格式不像我想要的那样。数据一次有4个不同玩家的playerID和playerTypes，如下所示：
dput(my.player.data)
structure(list(p_id = c(8470828L, 8478460L, 8470966L, 8475314L,
8476472L, 8476917L, 8475791L, 8470105L, 8476905L, 8474152L, 8470642L,
8479325L, 8475218L, 8471296L, 8476874L, 8477943L, 8477934L, 8473432L
), pType = c("Blocker", "Shooter", "Blocker", "Shooter", "Blocker",
"Hitter", "Blocker", "Shooter", "PlayerID", "PlayerID", "Shooter",
"Hitter", "PlayerID", "Blocker", "Shooter", "Scorer", "Scorer",
"Scorer"), p_id1 = c(8475172L, 8470645L, 8474162L, NA, 8480172L,
8477989L, 8476879L, NA, NA, NA, NA, 8474683L, NA, 8476851L, 8469514L,
8477407L, 8478402L, 8474091L), pType1 = c("Shooter", "Goalie",
"Shooter", NA, "Shooter", "Hittee", "Shooter", NA, NA, NA, NA,
"Hittee", NA, "Shooter", "Goalie", "Assist", "Assist", "Assist"
), p_id2 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, 8475246L, 8471729L, 8477018L), pType2 = c(NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "Assist",
"Assist", "Assist"), p_id3 = c(NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, 8475622L, 8471239L, 8469608L), pType3 = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "Goalie",
"Goalie", "Goalie")), .Names = c("p_id", "pType", "p_id1", "pType1",
"p_id2", "pType2", "p_id3", "pType3"), row.names = c(1L, 5001L,
10001L, 15001L, 20001L, 25001L, 30001L, 35001L, 40001L, 45001L,
50001L, 55001L, 60001L, 65001L, 70001L, 47329L, 46786L, 45551L
), class = "data.frame")
# ignore that the row numbers are 1, 5000, 10000, etc.
head(my.player.data)
p_id pType p_id1 pType1 p_id2 pType2 p_id3 pType3
1 8470828 Blocker 8475172 Shooter NA <NA> NA <NA>
5001 8478460 Shooter 8470645 Goalie NA <NA> NA <NA>
10001 8470966 Blocker 8474162 Shooter NA <NA> NA <NA>
15001 8475314 Shooter NA <NA> NA <NA> NA <NA>
20001 8476472 Blocker 8480172 Shooter NA <NA> NA <NA>
25001 8476917 Hitter 8477989 Hittee NA <NA> NA <NA>
我的数据中只有固定数量的pTypes在4个pType列中（Blocker，Shooter，Goalie等），我想为每个列创建一个列，列中的值等于相应的playerID。
例如，我想要一些看起来像这样的东西：
head(better.player.data)
Blocker Shooter Hittee Hitter Assist1 Assist2 Scorer Goalie
1 8470828 8475172 NA NA NA NA NA NA
5001 NA 8478460 NA NA NA NA NA 8470645
10001 8470966 8474162 NA NA NA NA NA NA
15001 NA 8475314 NA NA NA NA NA NA
20001 8476472 8480172 NA NA NA NA NA NA
25001 NA NA 8477989 8476917 NA NA NA NA
这里的主要边缘情况是Assist1和Assist2都在my.player.data数据框中标记为Assist（参见最后3行，未显示在head（）中）。我希望p_id1为Assist1，p_id2为Assist2（pType1和pType2应该是原始数据中唯一的2列，其值为Assist（不应该是pType或pType3）
我们非常感谢您对此的任何帮助！谢谢！
答案 0 :(得分：2)
使用tidyverse可以实现一个解决方案。我们的想法是gather以长格式转换，rowname具有匹配的pType和p_id列。首先根据与p_id和pType列相关联的数字创建一个组。执行修改以将Assist转换为Assist1和Assist2。最后，调用spread以转换所需格式的数据。
library(tidyverse)
my.player.data %>% rownames_to_column %>%
mutate(rowname = as.numeric(rowname)) %>%
gather(Key, Value, -rowname) %>%
filter(!is.na(Value)) %>%
mutate(Group = as.integer(gsub("(p_id|pType)","0",Key))) %>%
mutate(Value = ifelse(Value == "Assist", paste0(Value, Group), Value)) %>%
mutate(Key = gsub("\\d","",Key)) %>% #Remove number from p_id and pType columns
spread(Key, Value) %>%
select(-Group) %>%
spread(pType, p_id) %>%
remove_rownames() %>%
column_to_rownames()
# Assist1 Assist2 Blocker Goalie Hittee Hitter PlayerID Scorer Shooter
# 1 <NA> <NA> 8470828 <NA> <NA> <NA> <NA> <NA> 8475172
# 5001 <NA> <NA> <NA> 8470645 <NA> <NA> <NA> <NA> 8478460
# 10001 <NA> <NA> 8470966 <NA> <NA> <NA> <NA> <NA> 8474162
# 15001 <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> 8475314
# 20001 <NA> <NA> 8476472 <NA> <NA> <NA> <NA> <NA> 8480172
# 25001 <NA> <NA> <NA> <NA> 8477989 8476917 <NA> <NA> <NA>
# 30001 <NA> <NA> 8475791 <NA> <NA> <NA> <NA> <NA> 8476879
# 35001 <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> 8470105
# 40001 <NA> <NA> <NA> <NA> <NA> <NA> 8476905 <NA> <NA>
# 45001 <NA> <NA> <NA> <NA> <NA> <NA> 8474152 <NA> <NA>
# 45551 8474091 8477018 <NA> 8469608 <NA> <NA> <NA> 8473432 <NA>
# 46786 8478402 8471729 <NA> 8471239 <NA> <NA> <NA> 8477934 <NA>
# 47329 8477407 8475246 <NA> 8475622 <NA> <NA> <NA> 8477943 <NA>
# 50001 <NA> <NA> <NA> <NA> <NA> <NA> <NA> <NA> 8470642
# 55001 <NA> <NA> <NA> <NA> 8474683 8479325 <NA> <NA> <NA>
# 60001 <NA> <NA> <NA> <NA> <NA> <NA> 8475218 <NA> <NA>
# 65001 <NA> <NA> 8471296 <NA> <NA> <NA> <NA> <NA> 8476851
# 70001 <NA> <NA> <NA> 8469514 <NA> <NA> <NA> <NA> 8476874
答案 1 :(得分：2)
library(data.table)
dcast(na.omit(
melt(setDT(df)[,id:=1:nrow(df)],"id",list(grep("p_id",names(df)),grep("pType",names(df))))
),
id~value2,value.var = "value1")
id Assist1 Assist2 Blocker Goalie Hittee Hitter PlayerID Scorer Shooter
1: 1 NA NA 8470828 NA NA NA NA NA 8475172
2: 2 NA NA NA 8470645 NA NA NA NA 8478460
3: 3 NA NA 8470966 NA NA NA NA NA 8474162
4: 4 NA NA NA NA NA NA NA NA 8475314
5: 5 NA NA 8476472 NA NA NA NA NA 8480172
6: 6 NA NA NA NA 8477989 8476917 NA NA NA
7: 7 NA NA 8475791 NA NA NA NA NA 8476879
8: 8 NA NA NA NA NA NA NA NA 8470105
9: 9 NA NA NA NA NA NA 8476905 NA NA
10: 10 NA NA NA NA NA NA 8474152 NA NA
11: 11 NA NA NA NA NA NA NA NA 8470642
12: 12 NA NA NA NA 8474683 8479325 NA NA NA
13: 13 NA NA NA NA NA NA 8475218 NA NA
14: 14 NA NA 8471296 NA NA NA NA NA 8476851
15: 15 NA NA NA 8469514 NA NA NA NA 8476874
16: 16 8477407 8475246 NA 8475622 NA NA NA 8477943 NA
17: 17 8478402 8471729 NA 8471239 NA NA NA 8477934 NA
18: 18 8474091 8477018 NA 8469608 NA NA NA 8473432 NA
如果你想要一个基础R方法： 你可以这样做：
as.data.frame.matrix(
xtabs(p_id~id+pType,
reshape(transform(df,pType1 =sub("(Assist)","\\11",pType1),pType2 = sub("(Assist)","\\12",pType2), id=1:nrow(df)),matrix(1:(ncol(a)-1),2),dir="long")),
row.names(df))
Assist1 Assist2 Blocker Goalie Hittee Hitter PlayerID Scorer Shooter
1 0 0 8470828 0 0 0 0 0 8475172
5001 0 0 0 8470645 0 0 0 0 8478460
10001 0 0 8470966 0 0 0 0 0 8474162
15001 0 0 0 0 0 0 0 0 8475314
20001 0 0 8476472 0 0 0 0 0 8480172
25001 0 0 0 0 8477989 8476917 0 0 0
30001 0 0 8475791 0 0 0 0 0 8476879
35001 0 0 0 0 0 0 0 0 8470105
40001 0 0 0 0 0 0 8476905 0 0
45001 0 0 0 0 0 0 8474152 0 0
50001 0 0 0 0 0 0 0 0 8470642
55001 0 0 0 0 8474683 8479325 0 0 0
60001 0 0 0 0 0 0 8475218 0 0
65001 0 0 8471296 0 0 0 0 0 8476851
70001 0 0 0 8469514 0 0 0 0 8476874
47329 8477407 8475246 0 8475622 0 0 0 8477943 0
46786 8478402 8471729 0 8471239 0 0 0 8477934 0
45551 8474091 8477018 0 8469608 0 0 0 8473432 0
答案 2 :(得分：1)
在进行一些预处理之后，我们可以将tidyverse::spread与reduce一起使用，因为同一行可以包含Assist的多个值：
library(tidyverse)
df %>%
rownames_to_column %>%
mutate(pType1 = gsub("Assist","Assist1",pType1),
pType2 = gsub("Assist","Assist2",pType2)) %>%
reduce(.init= .,.x=1:4,~spread(.,3,2))
# rowname Blocker Hitter PlayerID Scorer Assist1 Hittee Shooter Assist2 Goalie <NA>
# 1 1 8470828 NA NA NA NA NA 8475172 NA NA NA
# 2 10001 8470966 NA NA NA NA NA 8474162 NA NA NA
# 3 15001 NA NA NA NA NA NA NA NA NA NA
# 4 20001 8476472 NA NA NA NA NA 8480172 NA NA NA
# 5 25001 NA 8476917 NA NA NA 8477989 NA NA NA NA
# 6 30001 8475791 NA NA NA NA NA 8476879 NA NA NA
# 7 35001 NA NA NA NA NA NA NA NA NA NA
# 8 40001 NA NA 8476905 NA NA NA NA NA NA NA
# 9 45001 NA NA 8474152 NA NA NA NA NA NA NA
# 10 45551 NA NA NA 8473432 8474091 NA NA 8477018 8469608 NA
# 11 46786 NA NA NA 8477934 8478402 NA NA 8471729 8471239 NA
# 12 47329 NA NA NA 8477943 8477407 NA NA 8475246 8475622 NA
# 13 50001 NA NA NA NA NA NA NA NA NA NA
# 14 5001 NA NA NA NA NA NA NA NA NA NA
# 15 55001 NA 8479325 NA NA NA 8474683 NA NA NA NA
# 16 60001 NA NA 8475218 NA NA NA NA NA NA NA
# 17 65001 8471296 NA NA NA NA NA 8476851 NA NA NA
# 18 70001 NA NA NA NA NA NA NA NA NA NA
答案 3 :(得分：1)
这是一个基础R版本：
gty <- function(.) unlist(.[grep("pType", names(.))])
# a function to get the vector of "types" from a row of the data frame
# edit: I noticed the "Assist" problem ("Assist1" and "Assist2" in the target table but "Assist" in the source table)
# ... so this needs to be corrected
gty <- function(.) {
res <- unlist(.[grep("pType", names(.))])
res[res %in% "Assist"] <- paste0("Assist", 1:sum(res%in%"Assist"))
res
}
# or in a hyper-functional style
gty <- function(.){ (function(..) {"[<-" (.., ..%in%"Assist", paste0("Assist", 1:sum(..%in%"Assist")))})(unlist(.[grep("pType", names(.))]))}
gid <- function(.) as.numeric(unlist(.[grep("_id", names(.))]))
# a function to get the vector of "ids" from a row of the data frame
# as.numeric - needed if this will be used from within `apply`as I intend to
types <- c("Blocker", "Shooter", "Hittee", "Hitter", "Assist1", "Assist2",
"Scorer", "Goalie")
# a list of types
fun <- function(.) setNames(gid(.)[match(types, gty(.))], types)
# a function which gets the ids and types from a row and rearranges them
t(apply(df, 1, fun))
结果：
Blocker Shooter Hittee Hitter Assist1 Assist2 Scorer Goalie
1 8470828 8475172 NA NA NA NA NA NA
5001 NA 8478460 NA NA NA NA NA 8470645
10001 8470966 8474162 NA NA NA NA NA NA
15001 NA 8475314 NA NA NA NA NA NA
20001 8476472 8480172 NA NA NA NA NA NA
25001 NA NA 8477989 8476917 NA NA NA NA
30001 8475791 8476879 NA NA NA NA NA NA
35001 NA 8470105 NA NA NA NA NA NA
40001 NA NA NA NA NA NA NA NA
45001 NA NA NA NA NA NA NA NA
50001 NA 8470642 NA NA NA NA NA NA
55001 NA NA 8474683 8479325 NA NA NA NA
60001 NA NA NA NA NA NA NA NA
65001 8471296 8476851 NA NA NA NA NA NA
70001 NA 8476874 NA NA NA NA NA 8469514
47329 NA NA NA NA 8477407 8475246 8477943 8475622
46786 NA NA NA NA 8478402 8471729 8477934 8471239
45551 NA NA NA NA 8474091 8477018 8473432 8469608
答案 4 :(得分：0)
不是我想要的创造性答案，但这在技术上对我有用：
my.player.data %>%
dplyr::mutate(Shooter = ifelse(pType == "Shooter", p_id, ifelse(pType1 == "Shooter", p_id1, ifelse(pType2 == "Shooter", p_id2, ifelse(pType3 == "Shooter", p_id3, NA))))) %>%
dplyr::mutate(Goalie = ifelse(pType == "Goalie", p_id, ifelse(pType1 == "Goalie", p_id1, ifelse(pType2 == "Goalie", p_id2, ifelse(pType3 == "Goalie", p_id3, NA))))) %>%
dplyr::mutate(Blocker = ifelse(pType == "Blocker", p_id, ifelse(pType1 == "Blocker", p_id1, ifelse(pType2 == "Blocker", p_id2, ifelse(pType3 == "Blocker", p_id3, NA)))))
dplyr::mutate(Assister1 = ifelse(pType1 == "Assist", p_id1, NA)) %>%
dplyr::mutate(Assister1 = ifelse(pType2 == "Assist", p_id2, NA))