Question

我有一个名为df的数据：（没有重复的df行）

a_id           b_id

111111         18
111111         17
222222         18
333333         14
444444         13 
555555         18
555555         24
222222         13
222222         17
333333         17

我希望将其反转为df_2这样的数据：

a_one     a_two      b_list   number_of_b  
222222    444444     13       1
111111    222222     17,18    2
111111    333333     17       1
111111    222222     17       1
222222    333333     17       1
111111    555555     18       1
222222    555555     18       1

如果a_id共享相同的b_id，则它们在df_2;

上成为一对

df_2的b_list是相应的b_id;

number_of_b是b_list

的长度

我有一个python代码

import pandas as pd
from itertools import combinations
df = df.groupby("b_id").apply(lambda x: list(combinations(x["a_id"], 2))).apply(pd.Series).stack()
df = df.apply(pd.Series).reset_index().groupby([0,1])["b_id"].apply(lambda x:x.values).reset_index()
df.columns = ["a_one", "a_two", "b_list"]
df["number_of_b"] = df.b_list.apply(len)

任何人都可以帮我在R

中实现它

Answer 1

我的方法有点长，但会给你想要的结果。这是我的方法：

library(data.table)
mydf <- data.table(structure(list(a_id = c(111111L, 111111L, 222222L, 333333L, 444444L, 
                                           555555L, 555555L, 222222L, 222222L, 333333L), b_id = c(18L, 17L, 
                                                                                                  18L, 14L, 13L, 18L, 24L, 13L, 17L, 17L)), .Names = c("a_id", 
                                                                                                                                                       "b_id"), class = "data.frame", row.names = c(NA, -10L)))

mydf <- mydf[mydf,.(a_id,a_id2=i.a_id,b_id),on="b_id",allow.cartesian=TRUE][a_id!=a_id2]

# Find duplicates
get_index <- function(string,values,current_index){
  string_present <- match(string,values)
  string_present[string_present<current_index] <- 0
  return(string_present)
}

mydf[,c("first","reverse"):= .(paste0(a_id,", ",a_id2,", ",b_id),paste0(a_id2,", ",a_id,", ",b_id))]
mydf[,duplicate_index:= get_index(first,reverse,.I)]
mydf[duplicate_index==0,.(b_list=list(b_id),number_of_b=.N),.(a_id,a_id2)]
# a_id  a_id2 b_list number_of_b
# 1: 111111 222222  18,17           2
# 2: 111111 555555     18           1
# 3: 222222 555555     18           1
# 4: 444444 222222     13           1
# 5: 111111 333333     17           1
# 6: 222222 333333     17           1

Answer 2

使用base-R ......

df2 <- tapply(df$a_id, df$b_id, sort) #gather sorted a ids by b
df2 <- df2[sapply(df2, function(x) length(x)>1)] #remove single items
df2 <- stack(lapply(df2, function(x) apply(combn(x,2), 2, paste, collapse=" "))) #paste a's in pairs
df2 <- as.data.frame(tapply(df2$ind, df2$values, paste, collapse=",")) #gather b ids by a pairs
names(df2) <- "b_list" 
df2[,c("a_one","a_two")] <- do.call(rbind,strsplit(rownames(df2)," ")) #create a columns from row names
df2$number_of_b <- sapply(df2$b_list,function(x) length(strsplit(x,",")[[1]]))
rownames(df2) <- NULL #remove row names
df2 <- df2[,c(2,3,1,4)] #reorder columns

df2
   a_one  a_two b_list number_of_b
1 111111 222222  17,18           2
2 111111 333333     17           1
3 111111 555555     18           1
4 222222 333333     17           1
5 222222 444444     13           1
6 222222 555555     18           1

Answer 3

合并到自身，对列进行排序并删除重复项，然后总结：

library(dplyr)

merge(df1, df1, by = "b_id") %>% 
  transmute(a_one = pmin(a_id.x, a_id.y),
            a_two = pmax(a_id.x, a_id.y),
            b_id) %>% 
  filter(a_one != a_two) %>% 
  unique() %>% 
  group_by(a_one, a_two) %>% 
  summarise(b_list = paste(b_id, collapse = ","),
            number_of_b = n())

# # A tibble: 6 x 4
# # Groups:   a_one [?]
#    a_one  a_two b_list number_of_b
#    <int>  <int>  <chr>       <int>
# 1 111111 222222  17,18           2
# 2 111111 333333     17           1
# 3 111111 555555     18           1
# 4 222222 333333     17           1
# 5 222222 444444     13           1
# 6 222222 555555     18           1

R：按照一定的模式传输数据表

3 个答案: