当行中存在与另一行相同的值时为行分配标签

Question

I am doing some data cleaning and I came across this problem. An example of my dataframe (some entries are missing) is:

A	B	C
1	7
1		8
1		9
1
2	2	5
2		5
		3
4	5	9
	5

and my expected output dataframe is:

A	B	C	Label
1	7		0
1		8	0
1		9	0
1			0
2	2	5	1
2		5	1
		3	2
4	5	9	0
	5		0

Are there ways in pandas/dplyr to get this output?

Edit: For example, the value 1 appears in the first 4 rows of A so these rows should have the same label. The 3rd row and 2nd last row has value 9 in column C so they should also have the same label. The last 2 rows have values 5 in column B so they should also be the same label. The 3rd last row does not have any values in the row that matches any value in the same column of any row so it is a unique label.

Answer 1

This will accomplish the task. In your example, there is a slight conflict with your comment. Since rows 8 and 9 have label 0 and contain 5's in at least one column, rows 5 and 6 should retroactively be changed to label 0, because they contain 5's in at least one column as well.

text="A B   C
1   7   NA
1   NA  8
1   NA  9
1   NA NA   
2   2   5
2   NA  5
3 NA NA
4   5   9
5 NA NA"
df=read.table(text=text, header=TRUE)
my_maps=list()
my_maps[[1]]=unique(unlist(df[1,]))
my_maps[[1]]=my_maps[[1]][!is.na(my_maps[[1]])]
for (i in 2:nrow(df)) {
  currow=df[i,]
  currow=unique(unlist(currow))
  currow=currow[!is.na(currow)]
  boo=TRUE
  for (k in 1:length(my_maps)) {
    if (length(intersect(my_maps[[k]], currow))>0) {
      my_maps[[k]]=union(my_maps[[k]], currow)
      boo=FALSE
    }
  }
  if (boo) {
    my_maps[[length(my_maps)+1]]=currow
  }
}

i=1
while (i < length(my_maps)) {
  j=i+1
  while (j <= length(my_maps)) {
    if (length(intersect(my_maps[[i]], my_maps[[j]]))>0) {
      my_maps[[i]]=union(my_maps[[i]], my_maps[[j]])
      my_maps[[j]]=NULL
    }
    j=j+1
  }
  i=i+1
}

label=c()
for (i in 1:nrow(df)) {
  for (j in 1:length(my_maps)) {
    if (length((intersect(my_maps[[j]], unique(unlist(df[i,])))))>0) {
      label[i]=j
      break
    }
  }
}

df=mutate(df, label=label-1)

  A  B  C label
1 1  7 NA     0
2 1 NA  8     0
3 1 NA  9     0
4 1 NA NA     0
5 2  2  5     0
6 2 NA  5     0
7 3 NA NA     1
8 4  5  9     0
9 5 NA NA     0