I have a DF with 800k+ rows with repeated (random) values. For each row I need to take a value and find an index of a new row(s) with same value. E.g. "asd" - where else do I see it? The index of the current row is NOT needed.
My current solution: subset a DF and create a temp frame/table by removing current row. Problem - it takes a minute per 1000 iterations. So 800+k rows will take me 13 hours to run. Any ideas? Thanks!
Running on original DF (not subsetted) is < 1 second, but as you can imagine it gives me the index of the current row.
Edit: My real-life DF is more than 1 column. Example below is simplified. I need to take V1[1]
and get row numbers of other V1
with value of V1[1]
, then repeat for V1[2]
and so on for each row
library(fastmatch)
library(stringi)
set.seed(12345)
V1 = stringi::stri_rand_strings(800000, 3)
df0 = as.data.table(V1)
mapped = matrix("",nrow=800000)
print(Sys.time())
for (i in 1:1000) {
tmp_df = df0[-i,] #This takes very long time!!!
mapped[i] = fmatch(df0$V1[i],tmp_df$V1)
}
print(Sys.time())
View(mapped)
Data:
library("data.table")
set.seed(12345)
V1 = stringi::stri_rand_strings(80, 3)
df0 <- data.table( sample(V1, 100, replace = TRUE ))
Code:
df0[, id := list(list(.I)), by = V1] # integer id
Output:
head(df0, 10)
# V1 id
# 1: iuR 1,2,21
# 2: iuR 1,2,21
# 3: KXc 3
# 4: LwA 4
# 5: pYn 5
# 6: qoN 6,66
# 7: 5Xt 7
# 8: wBH 8,77
# 9: V9r 9,39,54
# 10: 9ks 10,28,42,48
EDIT - Removed Current Index:
df0[, id2 := 1:.N ]
df0[, id := list(list(unlist(id)[ unlist(id) != .I ] )), by = id2 ]
df0[, id2 := NULL ]
df0[ lengths(id) > 0, ]
head( df0, 10 )
# V1 id
# 1: iuR 2,21
# 2: iuR 1,21
# 3: KXc
# 4: LwA
# 5: pYn
# 6: qoN 66
# 7: 5Xt
# 8: wBH 77
# 9: V9r 39,54
# 10: 9ks 28,42,48