我需要总结一下我分配给组的字符串数量,我知道我可以在dplyr/tidyr中完成,但我遗漏了一些东西.
示例数据集:
Owner = c('bob','julia','cheryl','bob','julia','cheryl')
Day = c('Mon', 'Tue')
Locn = c('house','store','apartment','office','house','shop')
data <- data.frame(Owner, Day, Locn)
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看起来像这样:
Owner Day Locn
1 bob Mon house
2 julia Tue store
3 cheryl Mon apartment
4 bob Tue office
5 julia Mon house
6 cheryl Tue shop
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我想按名称和日期分组,然后按列计算分组位置.在这个例子中,我希望"house"和"apartment"添加到标题为"Home"的列中,"store","office"和"shop"将计入"Work"列中.
我当前的代码(不起作用):
grouped_locn <- data %>%
dplyr::arrange(Owner, Day) %>%
dplyr::group_by(Owner, Day) %>%
dplyr::summarize(Home = which(data$Locn %in% c('house', 'apartment')),
Work = which(data$Locn %in% c("store", "office", "apartment")))
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我只是将我目前的尝试包括在总结步骤中,以显示我是如何接近它的.Home和Work代码当前返回包含组元素的行号的向量(即Home = 1 3 5)
我的预期输出:
Owner Day Home Work
1 bob Mon 1 0
2 bob Tue 0 1
3 julia Mon 1 0
4 julia Tue 0 1
5 cheryl Mon 1 0
6 cheryl Tue 0 1
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在实际数据集(30k +行)中,每个所有者每天有多个Locn值,因此Home和Work计数可以是1和0以外的数字(因此没有布尔值).
非常感谢.
Dav*_*urg 10
这是一个简单有效的解决方案 data.table
对于旧版本(v <1.9.5)
library(data.table) # v < 1.9.5
setDT(data)[, Locn2 := c("Work", "Home")[(Locn %in% c('house', 'apartment')) + 1L]]
dcast.data.table(data, Owner + Day ~ Locn2, length)
# Owner Day Home Work
# 1: bob Mon 1 0
# 2: bob Tue 0 1
# 3: cheryl Mon 1 0
# 4: cheryl Tue 0 1
# 5: julia Mon 1 0
# 6: julia Tue 0 1
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对于较新的版本(v> = 1.9.5),您可以在一行中执行此操作
dcast(setDT(data), Owner + Day ~ c("Work", "Home")[(Locn %in% c('house', 'apartment')) + 1L], length)
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这是tidyr另一种选择
library(dplyr)
library(tidyr)
data %>%
mutate(temp = 1L,
Locn = ifelse(Locn %in% c('house', 'apartment'), "Home", "Work")) %>%
spread(Locn, temp, fill = 0L)
# Owner Day Home Work
# 1 bob Mon 1 0
# 2 bob Tue 0 1
# 3 cheryl Mon 1 0
# 4 cheryl Tue 0 1
# 5 julia Mon 1 0
# 6 julia Tue 0 1
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试试这个
data %>%
group_by(Owner, Day) %>%
summarise(Home = sum(Locn %in% c("house", "apartment")),
Work = sum(Locn %in% c("store", "office", "shop")))
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