我想按位置名称汇总df。数据看起来像这样：
location <- c("NY", "NC", "KA", "TX", "AZ", "NC", "SC", "ND", "SD", "MN","WA","MA","VT","CA","OR","NJ","OH","MI","IL","GA","FL")
tree_type <- c("pine", "birch", "maple", "palm")
df <- data.frame(location = sample(location, 20, replace = TRUE),
tree_type = sample(tree_type, 20, replace = TRUE),
density = runif(20, min = 24, max = 365),
income = runif(20, min = 37000, max = 62000))
我想要的是：
location mean(density) mean(income) birch maple palm pine
1 AZ 38.44009 52032.95 0 0 1 0
2 CA 136.85112 42243.35 0 1 0 0
3 GA 101.24081 53405.60 2 0 0 0
4 IL 172.02651 46368.42 1 1 0 0
5 MA 198.69868 51117.18 0 0 0 1
6 MI 153.93358 60425.87 1 0 0 0
7 MN 185.05276 46468.68 0 0 1 0
8 NC 181.42187 46007.93 1 0 2 0
9 NJ 302.66541 59316.94 0 0 2 0
10 OR 303.88283 48497.03 0 0 0 2
11 SC 84.05136 50348.41 0 1 0 1
12 SD 158.47423 57894.27 0 0 1 0
13 VT 126.32967 42853.04 0 0 1 0
我是这样做的：
require(dplyr)
require(reshape2)
df_quantvars <- df %>% group_by(location) %>% summarise(mean(density), mean(income))
df_catvarslong <- as.data.frame(table(df[1:2]))
df_catvarswide <- dcast(df_catvarslong, location ~ tree_type, value.var = "Freq")
final_df <- left_join(df_quantvars, df_catvarswide, by = "location")
在dplyr group_by成语中是否无法做到这一点？冒着愚蠢的风险，我试着这样做：
df_quantvars <- df %>% group_by(location) %>% summarise(mean(density), mean(income), table(df[1:2]))
我错过了什么？
答案 0 :(得分：2)
这个回应有点迟，但我已经做了一些工作。保持一切只是有点棘手。这似乎有效：
首先我使用group_by(location, tree_type)来计算所有树，然后我使用group_by(location)来获得所需的方法。然后，我删除原始密度＆amp;带有select(-c(density, income)的收入类别，并且留有重复的行但是正确的聚合计数。然后，我使用distinct()删除重复项，然后使用spread()库中的tidyr转换为您请求的宽格式。
library(dplyr)
library(tidyr)
df %>%
arrange(location)%>%
group_by(location, tree_type)%>%
mutate(Count = n())%>%
group_by(location)%>%
mutate(MeanDensity = mean(density),
MeanIncome = mean(income))%>%
ungroup()%>%
select(-c(density, income))%>%
distinct()%>%
spread(key = tree_type, value = Count, fill = 0)
这给了我：
location MeanDensity MeanIncome birch maple palm pine
(fctr) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
1 AZ 244.18094 57474.94 0 0 1 0
2 FL 51.90693 42425.36 0 0 0 1
3 GA 341.18643 49385.44 0 0 0 2
4 IL 258.11124 37101.36 0 1 0 0
5 KA 267.92430 59699.20 1 0 0 0
6 MA 87.48623 60632.98 1 0 0 0
7 MI 197.18310 58837.00 0 0 0 1
8 NC 362.48531 50857.42 0 0 1 0
9 ND 315.57415 51465.06 0 0 1 0
10 NJ 233.72886 55877.40 0 0 1 1
11 NY 283.41522 49275.58 0 1 0 1
12 OH 350.23362 40901.73 0 0 1 0
13 OR 267.68415 38954.04 0 2 0 0
14 SC 260.12169 52837.10 0 1 0 0
15 SD 76.29782 54986.76 0 1 0 0
16 VT 341.80646 44547.77 1 0 0 0
答案 1 :(得分：0)
这可以分阶段进行。首先，找到一切手段。
x <- df %>%
group_by(location) %>%
summarise(mean(density), mean(income))
location `mean(density)` `mean(income)`
<fct> <dbl> <dbl>
1 AZ 150. 44667.
2 FL 262. 53719.
3 IL 308. 41077.
4 KA 183. 48432.
5 MI 192. 61649.
6 NC 210. 50838.
7 NJ 223. 49958.
8 OH 357. 57881.
9 OR 336. 52582.
10 SC 234. 40414.
11 SD 54.8 48203.
分别计算所有树类型，然后传播。
y <- df %>% group_by(location, tree_type) %>%
count() %>%
spread(tree_type, n, fill = 0L) #the fill option is great for replacing NAs with 0s
# A tibble: 11 x 5
# Groups: location [11]
location birch maple palm pine
<fct> <int> <int> <int> <int>
1 AZ 0 1 0 1
2 FL 0 0 1 0
3 IL 0 0 0 1
4 KA 2 0 0 2
5 MI 0 0 1 0
6 NC 1 2 0 1
7 NJ 1 0 1 0
8 OH 0 0 1 0
9 OR 1 0 0 0
10 SC 1 0 0 0
11 SD 1 0 1 0
最后，将两个数据框连接在一起。他们将根据各自共享的共同位置值自动加入。我使用了右连接，但是您也可以使用left_join。
right_join(x, y)
Joining, by = "location"
# A tibble: 11 x 7
location `mean(density)` `mean(income)` birch maple palm pine
<fct> <dbl> <dbl> <int> <int> <int> <int>
1 AZ 150. 44667. 0 1 0 1
2 FL 262. 53719. 0 0 1 0
3 IL 308. 41077. 0 0 0 1
4 KA 183. 48432. 2 0 0 2
5 MI 192. 61649. 0 0 1 0
6 NC 210. 50838. 1 2 0 1
7 NJ 223. 49958. 1 0 1 0
8 OH 357. 57881. 0 0 1 0
9 OR 336. 52582. 1 0 0 0
10 SC 234. 40414. 1 0 0 0
11 SD 54.8 48203. 1 0 1 0
三个阶段似乎很冗长。我们可以将所有内容组合成一个语句。
# This part is equivalent to stage one
df %>%
group_by(location) %>%
summarise(mean(density), mean(income)) %>%
# This part is equivalent to stage two and three.
right_join(
df %>% group_by(location, tree_type) %>%
count() %>%
spread(tree_type, n, fill = 0L)
)