I have a data table which contains multiple columns of the following type:
attr1 attr2
1: 01001 01000
2: 11000 10000
3: 00100 00100
4: 01100 01000
DT = setDT(structure(list(attr1 = c("01001", "11000", "00100", "01100"),
attr2 = c("01000", "10000", "00100", "01000")), .Names = c("attr1",
"attr2"), row.names = c(NA, -4L), class = "data.frame"))
All of the columns are strings not numbers. What I would like to achieve is the following:
1) I want to find the positions that the "1" appears in the strings of attr1
2) Take the values of attr2 in these positions
My result in this case would be:
[1] "10" "10" "1" "10"
As an example in the first row, attr1 has "1" in positions 2 and 5, I subset the first row of attr2 in positions 2 and 5 and end up with "10".
What I have thought to do is to strsplit the columns and then work with that but I really hope there is a better way.
You can use a variation on @alistaire's regmatches
answer, as there is also a replacement function regmatches<-
. So, instead of extracting 1
values, replace 0
values with ""
:
dt[, matches := `regmatches<-`(attr2, gregexpr("0+", attr1), value="")]
# attr1 attr2 matches
#1: 01001 01000 10
#2: 11000 10000 10
#3: 00100 00100 1
#4: 01100 01000 10
Your idea to strsplit
and compare is also feasible:
dt[, matches := mapply(function(x,y) paste(y[x==1],collapse=""), strsplit(attr1,""), strsplit(attr2,""))]
You can use base R's regmatches
to supply a different string for matching and replacement:
dt[, matches := sapply(regmatches(attr2, gregexpr('1+', attr1)), paste, collapse = '')][]
#> attr1 attr2 matches
#> 1: 01001 01000 10
#> 2: 11000 10000 10
#> 3: 00100 00100 1
#> 4: 01100 01000 10
Data
dt <- structure(list(attr1 = c("01001", "11000", "00100", "01100"),
attr2 = c("01000", "10000", "00100", "01000")), .Names = c("attr1",
"attr2"), row.names = c(NA, -4L), class = "data.frame")
setDT(dt)
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