Cpt*_*emo 3 for-loop r split-apply-combine dplyr data.table
我有97M行的长表.每行包含一个人采取的操作的信息以及该操作的时间戳,格式如下:
actions <- c("walk","sleep", "run","eat")
people <- c("John","Paul","Ringo","George")
timespan <- seq(1000,2000,1)
set.seed(28100)
df.in <- data.frame(who = sample(people, 10, replace=TRUE),
what = sample(actions, 10, replace=TRUE),
when = sample(timespan, 10, replace=TRUE))
df.in
# who what when
# 1 Paul eat 1834
# 2 Paul sleep 1295
# 3 Paul eat 1312
# 4 Ringo eat 1635
# 5 John sleep 1424
# 6 George run 1092
# 7 Paul walk 1849
# 8 John run 1854
# 9 George sleep 1036
# 10 Ringo walk 1823
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每个动作都可以由人采取或不采取,并且可以以任何顺序采取行动.
我有兴趣总结我的数据集中的操作顺序.特别是对于每个人,我想找到哪个动作是第一,第二,第三和第四.如果多次采取行动,我只对第一次出现感兴趣.那么,如果有人运行,吃,吃,跑和睡觉我感兴趣的总结,例如run,eat,sleep.
df.out <- data.frame(who = factor(character(), levels=people),
action1 = factor(character(), levels=actions),
action2 = factor(character(), levels=actions),
action3 = factor(character(), levels=actions),
action4 = factor(character(), levels=actions))
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我可以通过forloop获得我想要的东西:
for (person in people) {
tmp <- subset(df.in, who==person)
tmp <- tmp[order(tmp$when),]
chrono_list <- unique(tmp$what)
df.out <- rbind(df.out, data.frame(who = person,
action1 = chrono_list[1],
action2 = chrono_list[2],
action3 = chrono_list[3],
action4 = chrono_list[4]))
}
df.out
# who action1 action2 action3 action4
# 1 John sleep run <NA> <NA>
# 2 Paul sleep eat walk <NA>
# 3 Ringo eat walk <NA> <NA>
# 4 George sleep run <NA> <NA>
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这种结果是否也可以在没有循环的情况下以更有效的方式获得?
我们可以使用dcastdevel版本data.table,即.v1.9.5.我们可以安装它here
library(data.table)#v1.9.5+
dcast(setDT(df.in)[order(when),action:= paste0('action', 1:.N) ,who],
who~action, value.var='what')
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如果你需要unique每个'谁'的'什么'
dcast(setDT(df.in)[, .SD[!duplicated(what)], who][order(when),
action:= paste0('action', 1:.N), who], who~action, value.var='what')
# who action1 action2 action3
#1: George sleep run NA
#2: John sleep run NA
#3: Paul sleep eat walk
#4: Ringo eat walk NA
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或者使用.I会更快一些
ind <- setDT(df.in)[,.I[!duplicated(what)], who]$V1
dcast(df.in[ind][order(when),action:= paste0('action', 1:.N) ,who],
who~action, value.var='what')
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或者使用setorder,unique并且可以setorder通过引用对数据集重新排序,这可以是高效的内存.
dcast(unique(setorder(setDT(df.in), who, when), by=c('who', 'what'))[,
action:= paste0('action', 1:.N), who], who~action, value.var='what')
# who action1 action2 action3
#1: George sleep run NA
#2: John sleep run NA
#3: Paul sleep eat walk
#4: Ringo eat walk NA
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