我有一张长桌，上面有area和cluster的重复组合。
counts <- tibble::tribble(
~age, ~area, ~cluster, ~norm.to.area,
"gw_25", "cingulate", "cluster_1", 0.03,
"gw_20", "cingulate", "cluster_1", 0.03,
"gw_18", "hippocampus", "cluster_1", 0.02,
"gw_25", "insula", "cluster_1", 0.01,
"gw_20", "motor", "cluster_1", 0.01,
"gw_22", "motor", "cluster_1", 0.01,
"gw_25", "motor", "cluster_1", 0.01,
"gw_14", "motor", "cluster_1", 0.01,
"gw_18", "motor", "cluster_1", 0.01,
"gw_19", "motor", "cluster_1", 0.01,
"gw_17", "motor", "cluster_1", 0.01,
"gw_20", "occipital", "cluster_1", 0.01,
"gw_17", "occipital", "cluster_1", 0.01,
"gw_18", "occipital", "cluster_1", 0.01,
"gw_19", "occipital", "cluster_1", 0.01,
"gw_22", "occipital", "cluster_1", 0.01,
"gw_14", "occipital", "cluster_1", 0.01,
"gw_22", "parietal", "cluster_1", 0,
"gw_25", "parietal", "cluster_1", 0,
"gw_17", "parietal", "cluster_1", 0,
"gw_19", "parietal", "cluster_1", 0,
"gw_20", "parietal", "cluster_1", 0,
"gw_20", "PFC", "cluster_1", 0.01,
"gw_22", "PFC", "cluster_1", 0.01,
"gw_25", "PFC", "cluster_1", 0.01
)
我想创建一个新变量sum.norm.to.area，它是每个norm.to.area的{{1}}的总和，仅使用cluster的值对每个组合进行一次norm.to.area。
我尝试过area / subcluster.merge group_by，但这会根据给定组合出现的次数对这些值求和。
cluster
感谢您的建议。
更新1：
尝试使用汇总，如下所示，但是会发生同样的事情（当然，除了没有添加为新列）：
counts %>% group_by(cluster) %>% mutate(sum.norm.to.area = sum(norm.to.area)
> counts %>% group_by(subcluster.merge, area) %>% summarize(sum(norm.to.area))更新2
这是我想要的输出，但是我到达它的方式太复杂了。我想找到一种更简单的使用mutate的方式，而不必使用 tibble::tribble(
~cluster . , ~area, ~sum.norm.to.area.,
"cluster_1", "PFC", 0.06,
"cluster_1", "somatosensory", 0.05,
"cluster_1", "motor", 0.07,
"cluster_1", "parietal", 0,
"cluster_1", "temporal", 0.03,
"cluster_1", "occipital", 0.06,
"cluster_1", "hippocampus", 0.02,
"cluster_1", "insula", 0.01,
"cluster_1", "cingulate", 0.06,
"cluster_10-34", "PFC", 0.42,
"cluster_10-34", "somatosensory", 0.35,
"cluster_10-34", "motor", 0.48,
"cluster_10-34", "parietal", 0.36,
"cluster_10-34", "temporal", 0.28,
"cluster_10-34", "occipital", 0.4,
"cluster_10-34", "hippocampus", 0.12,
"cluster_10-34", "insula", 0,
"cluster_10-34", "cingulate", 0,
"cluster_11", "PFC", 0.18,
"cluster_11", "somatosensory", 0.15,
"cluster_11", "motor", 0.14,
"cluster_11", "parietal", 0.12,
"cluster_11", "temporal", 0.04,
"cluster_11", "occipital", 0.18,
"cluster_11", "hippocampus", 0.02
)
。
join所需的输出：
> tmp <- counts %>% distinct(area, cluster, .keep_all = TRUE) %>%
add_count(cluster, wt = norm.to.area, name = "sum.norm.to.area")
counts %>% left_join(tmp, by = c("cluster", "area"))
是为sum.norm.to.area和norm.to.area的所有唯一组合添加area（仅一次）的结果：
cluster答案 0 :(得分：1)
使用dplyr，我们可以group_by cluster和sum仅在每个area中使用唯一值。
library(dplyr)
counts %>%
group_by(cluster) %>%
mutate(sum.norm = sum(norm.to.area[!duplicated(area)]))
# age area cluster norm.to.area sum.norm
# <chr> <chr> <chr> <dbl> <dbl>
# 1 gw_25 cingulate cluster_1 0.03 0.09
# 2 gw_20 cingulate cluster_1 0.03 0.09
# 3 gw_18 hippocampus cluster_1 0.02 0.09
# 4 gw_25 insula cluster_1 0.01 0.09
# 5 gw_20 motor cluster_1 0.01 0.09
# 6 gw_22 motor cluster_1 0.01 0.09
# 7 gw_25 motor cluster_1 0.01 0.09
# 8 gw_14 motor cluster_1 0.01 0.09
# 9 gw_18 motor cluster_1 0.01 0.09
#10 gw_19 motor cluster_1 0.01 0.09
# … with 15 more rows
答案 1 :(得分：0)
我想您不是在寻找mutate（）
counts %>% group_by(cluster, area) %>% summarize(sum.norm.to.area = sum(norm.to.area))