Question

（此论坛上的其他问题和答案elsewhere似乎无法处理此Feed中提到的跨境问题）

假设我有以下数据：

df <- data.frame(id=c("Eric", "John", "Sarah", "Simon", "Abdul", "Charlotte", "Alex", "Susan"),
             state=c("CA", "AK", "NY", "NY", "NJ", "GA", "CA", "CA"),
             project=c(1, 2, 2, 2, 3, 4, 5, 5), stringsAsFactors = F)

> df
         id state project
1      Eric    CA       1
2      John    AK       2
3     Sarah    NY       2
4     Simon    NY       2
5     Abdul    NJ       3
6 Charlotte    GA       4
7      Alex    CA       5
8     Susan    CA       5

我想获得每个州的平均项目成员数量，也计算跨境成员。

为了获得只有州内成员的平均值，我做了以下几点：

dfx <- data.frame()
dfy <- data.frame()
for(j in unique(df$state)){
   h <- subset(df, state==j)
   counts <- plyr::count(h, 'project')
   #uniques <- length(unique(sub$invje))
   average_members <- mean(counts$freq)
   dfx <- data.frame(state=j,
                average_members=average_members)
   dfy <- rbind(dfy, dfx)
} 

> dfy
  state average_members
1    CA             1.5
2    AK             1.0
3    NY             2.0
4    NJ             1.0
5    GA             1.0

我想要的输出之后，AK和NY都应得3分，因为每个ID与项目中的其他两个ID一起工作（尽管生活在不同的状态）。

> desired
  state average_members
1    CA             1.5
2    AK             3.0
3    NY             3.0
4    NJ             1.0
5    GA             1.0

有谁知道如何编码？

Answer 1

library(data.table)

setDT(df)
df[, .(num_proj = .N), by = .(state, project)][, .(average_members = mean(num_proj)), by = state]

结果：

   state average_members
1:    CA             1.5
2:    AK             1.0
3:    NY             2.0
4:    NJ             1.0
5:    GA             1.0

对于第二种情况，请在第一次迭代中将state拉出群组。

unique(df[, .(state, num_proj = .N), by = project])[, .(average_members = mean(num_proj)), by = state]

1:    CA             1.5
2:    AK             3.0
3:    NY             3.0
4:    NJ             1.0
5:    GA             1.0

Answer 2

您可以使用dplyr库执行此操作。您可以使用

回答您在州内的唯一问题

library(dplyr)
df %>% count(state, project) %>% 
    group_by(state) %>% summarize(avg=mean(n))
#   state       avg
# 1    AK       1.0
# 2    CA       1.5
# 3    GA       1.0
# 4    NJ       1.0
# 5    NY       2.0

你可以用

获得跨州结果

df %>% distinct(project, state) %>% 
    inner_join(df %>% count(project)) %>% 
    group_by(state) %>% summarize(avg=mean(n))

#   state       avg
# 1    AK       3.0
# 2    CA       1.5
# 3    GA       1.0
# 4    NJ       1.0
# 5    NY       3.0

Answer 3

df <- data.frame(id=c("Eric", "John", "Sarah", "Simon", "Abdul", "Charlotte", "Alex", "Susan"),
                 state=c("CA", "AK", "NY", "NY", "NJ", "GA", "CA", "CA"),
                 project=c(1, 2, 2, 2, 3, 4, 5, 5), stringsAsFactors = F)


dfx <- data.frame()
dfy <- data.frame()

for (j in unique(df$state)) {
    h = subset(df, state==j)
    thisStatesProjects = unique(h[,"project"])
    h2 = subset(df, project %in% thisStatesProjects)
    average_members = nrow(h2)/length(thisStatesProjects)
    dfx <- data.frame(state=j,
                      average_members=average_members)
    dfy <- rbind(dfy, dfx)

}

R - 查找州内和州内的平均项目成员

3 个答案: