Question

我正在处理非常混乱的家庭数据，因为孩子们可能会与多个家庭进行分组。数据结构如下：

famid <- c("A","A","B","C","C","D","D")
kidid <- c("1","2","1","3","4","4","5")
df <- as.data.frame(cbind(famid, kidid))

我想根据这个家庭中所有孩子在另一个更大的家庭中聚集在一起的标准来确定我可以放弃哪些家庭。

例如，家庭A包含孩子1和孩子2.家庭B包含孩子1.因为家庭B完全包含在家庭A中，我想放弃家庭B.

或者，家庭C包含孩子3和孩子4.家庭D包含孩子4和孩子5.两个家庭都没有完全包含在另一个家庭中，所以我不想暂时放弃。

在我的数据中，每个孩子最多可以有6个家庭，每个家庭最多可以有8个孩子。有成千上万的家庭和成千上万的孩子。

我已经尝试通过创建一个非常宽的data.frame来解决这个问题，每个学生有一行，孩子们关联的每个家庭的列，孩子与之关联的每个家庭中的每个兄弟，以及一个额外的列（ sibgrp）用于将所有兄弟姐妹连接在一起的每个相关联的家庭。但是当我试图在连接字符串中搜索单个兄弟姐妹时，我发现我并不知道如何做到这一点 - grepl不会将矢量作为模式参数。

然后我开始研究相交和相似的函数，但是那些相互比较了整个矢量，而不是矢量内的观察到该矢量内的其他观测。（含义 - 我无法查找字符串df[1,2]和字符串df[1,3]之间的交叉点。相交会识别df[2]和df[3]之间的交叉点。

我试图改变我的想法以适应这种方法，这样我就可以比较兄弟姐妹彼此的向量，假设我已经知道至少有一个兄弟姐妹是共享的。考虑到有多少不同的家庭，我甚至无法弄清楚如何开始这样做，甚至一个共同的孩子也有多少不相关。

我在这里缺少什么？非常感谢任何反馈。谢谢！

Answer 1

此功能也可用于执行任务。它返回一个字符向量，其中包含可以删除的族的名称。

test_function <- function(dataset){

## split the kidid on the basis of famid
kids_family <- split.default(dataset[['kidid']],f = dataset[['famid']])

family <- names(kids_family)

## This function generates all the possible combinations if we select any two families from family
combn_family <- combn(family,2)

family_removed <- character(0)
apply(combn_family,MARGIN = 2, function(x){

  if (length(setdiff(kids_family[[x[1]]],kids_family[[x[2]]])) == 0)
    family_removed <<- c(family_removed,x[1])
  else if (length(setdiff(kids_family[[x[2]]],kids_family[[x[1]]])) == 0)
    family_removed <<- c(family_removed,x[2])

})

return (family_removed)
}
> df <- data.frame(famid = c("A","A","B","C","C","D","D", "E", "E", "E", "F", "F"),
+                  kidid = c(1, 2, 1, 3, 4, 4, 5, 7, 8, 9, 7, 9))
> test_function(df)
[1] "B" "F"

Answer 2

我没有机会试过setdiff。我来发布这个费力的解决方案，希望有更好的方法。

# dependencies for melting tables and handling data.frames
require(reshape2)
require(dplyr)


# I have added two more cases to your data.frame
# kidid is passed as numeric (with quoted would have been changed to vector by default)
df <- data.frame(famid = c("A","A","B","C","C","D","D", "E", "E", "E", "F", "F"),
                 kidid = c(1, 2, 1, 3, 4, 4, 5, 7, 8, 9, 7, 9))

# let's have a look to it
df
famid kidid
1      A     1
2      A     2
3      B     1
4      C     3
5      C     4
6      D     4
7      D     5
8      E     7
9      E     8
10     E     9
11     F     7
12     F     9

# we build a contingency table
m <- table(df$famid, df$kidid)

# a family A only contains a family B, if A has all the elements of B, 
# and at least one that B doesnt have
m

  1 2 3 4 5 7 8 9
A 1 1 0 0 0 0 0 0
B 1 0 0 0 0 0 0 0
C 0 0 1 1 0 0 0 0
D 0 0 0 1 1 0 0 0
E 0 0 0 0 0 1 1 1
F 0 0 0 0 0 1 0 1

# an helper function to implement that and return a friendly data.frame
family_contained <- function(m){
  res <- list()
  for (i in 1:nrow(m))
    # for each line in m, we calculate the difference to all other lines
    res[[i]] <- t(apply(m[-i, ], 1, function(row) m[i, ] - row))
  # here we test if all values are 0+ (ie if the selected family has all element of the other)
  # and if at least one is >=1 (ie if the selected family has at least one element that the other doesnt have)
  tab <- sapply(res, function(m) apply(m, 1,  function(x) all(x>=0) & any(x>=1)))
  # we format it as a table to have nice names
  tab %>% as.table() %>% 
    # we melt it into a data.frame
    melt()  %>% 
    # only select TRUE and get rid of this column
    filter(value) %>% select(-value) %>% 
    # to make things clear we name columns
    `colnames<-`(c("this_family_is_contained", "this_family_contains"))
}

family_contained(m)
# this_family_is_contained this_family_contains
# 1           B               A
# 2           F               E

# finally you can filter them with
filter(df, !(famid %in% family_contained(m)$this_family_is_contained))

根据2个向量中的信息确定重叠组的相对大小

2 个答案: