我在R中有一个数据帧df，就像这样。我想根据hee_provn1
的不同值循环显示dfnpi_one npi_two hee_provn1
1 2 175221
3 4 175221
5 6 175221
7 8 175221
9 10 576546
11 12 576546
13 14 576546
15 16 789535
17 18 789535
19 20 789535
现在我的R代码是：
library(dplyr)
library(igraph)
df2 <- filter(df, hee_provn1 == '175221')
df3 <- df2 [,c("npi_one","npi_two")]
l = c(apply(df3,1,c))
G <- graph(l,directed = FALSE )
degree(G) -> d
closeness(G) -> c
betweenness(G) -> b
eigen_centrality(G)$vector -> e
cent_df = data.frame(d,c,b,e)
colnames(cent_df) <- c('degree', 'closeness','betweenness','eigen')
cbind(hee_provn1 = 175221,cent_df)
第一个循环的结果表cent_df（hee_provn1 = 175221）是
hee_provn1 degree closeness betweenness eigen
1 175221 1 0.02040816 0 0.3227867
2 175221 1 0.02040816 0 0.3227867
3 175221 1 0.02040816 0 0.0000000
4 175221 1 0.02040816 0 0.0000000
5 175221 1 0.02040816 0 1.0000000
6 175221 1 0.02040816 0 1.0000000
7 175221 1 0.02040816 0 0.0000000
8 175221 1 0.02040816 0 0.0000000
第二个循环的结果表cent_df（hee_provn1 = 576546）是
hee_provn1 degree closeness betweenness eigen
1 576546 0 0.005494505 0 0
2 576546 0 0.005494505 0 0
3 576546 0 0.005494505 0 0
4 576546 0 0.005494505 0 0
5 576546 0 0.005494505 0 0
6 576546 0 0.005494505 0 0
7 576546 0 0.005494505 0 0
8 576546 0 0.005494505 0 0
9 576546 1 0.005917160 0 1
10 576546 1 0.005917160 0 1
11 576546 1 0.005917160 0 0
12 576546 1 0.005917160 0 0
13 576546 1 0.005917160 0 0
14 576546 1 0.005917160 0 0
我的想法结果是循环，我可以将所有结果表放在一个大表中，如
hee_provn1 degree closeness betweenness eigen
1 175221 1 0.02040816 0 0.3227867
2 175221 1 0.02040816 0 0.3227867
3 175221 1 0.02040816 0 0.0000000
4 175221 1 0.02040816 0 0.0000000
5 175221 1 0.02040816 0 1.0000000
6 175221 1 0.02040816 0 1.0000000
7 175221 1 0.02040816 0 0.0000000
8 175221 1 0.02040816 0 0.0000000
9 576546 0 0.005494505 0 0
10 576546 0 0.005494505 0 0
11 576546 0 0.005494505 0 0
12 576546 0 0.005494505 0 0
13 576546 0 0.005494505 0 0
14 576546 0 0.005494505 0 0
15 576546 0 0.005494505 0 0
16 576546 0 0.005494505 0 0
17 576546 1 0.005917160 0 1
18 576546 1 0.005917160 0 1
19 576546 1 0.005917160 0 0
20 576546 1 0.005917160 0 0
21 576546 1 0.005917160 0 0
22 576546 1 0.005917160 0 0
我真的希望它尽可能高效。
答案 0 :(得分：1)
您的示例数据
df <- data.frame(npi_one=seq(1,19,2),
npi_two=seq(2,20,2),
hee_provn1=c(rep(175221,4),rep(576546,3),rep(789535,3)))
除igraph外，您还需要tidyverse
library(tidyverse)
library(igraph)
我已注释以下代码以匹配原始代码的代码
final <- df %>%
group_by(hee_provn1) %>% # similar to filter(df, hee_provn1 == '175221')
nest() %>% # similar to df2 [,c("npi_one","npi_two")]
mutate(data=map(data,~c(apply(.x,1,c)))) %>% # similar to c(apply(df3,1,c))
mutate(data=map(data,~graph(.x,directed=F))) %>% # similar to graph(l,directed = FALSE )
mutate(data=map(data,~ data.frame( degree = degree(.x),
closeness = closeness(.x),
betweenness = betweenness(.x),
eigen_centrality = eigen_centrality(.x)$vector ) ) ) %>% # similar to making b, c, d, e individually
unnest(data) # revert to normal data frame
输出head(final)
hee_provn1 degree closeness betweenness eigen_centrality
1 175221 1 0.020408163 0 1.000000e+00
2 175221 1 0.020408163 0 1.000000e+00
3 175221 1 0.020408163 0 0.000000e+00
4 175221 1 0.020408163 0 0.000000e+00
注意每次运行eigen_centrality时，我都会获得不同的值，因此请确保返回您期望的值