Data checking function to check if the contents of two variables have a 1:1 mapping

isOneToOne(x, y)

checkIsOneToOne(x, y)

Arguments

x

first variable

y

second variable

Value

logical: TRUE if 1:1 mapping, FALSE otherwise

Background

This is a little utility function that I use from time to time when I want to check whether two variables have a one to one mapping they should have. Typically this is where one of them is a short code say variable x, for the other, say variable y. If you have two vectors, one of the short codes, x, and one of the longer, y, they should have a perfect 1:1 mapping, so for any value in x there must the same number of its mapped value in y. For example:

xy
1Male
2Female
3Other
2Female
2Female
1Male
3Other
Has a perfect 1:1 mapping of x to y so, assuming that
I had x and y as vectors,
isOneToOne(x, y) will return TRUE.
xy
1Male
2Feemale
3Other
2Female
2Female
1male
3Other
does not 1:1 (and isn't a completely mad example
of typical messes in real world data, if perhaps a very
simple one!) Of course, here
isOneToOne(x, y) will return FALSE.

Acknowledgements

I found this nice way of doing this at https://stackoverflow.com/questions/52399474/check-if-variables-are-in-a-one-to-one-mapping

History/development log

Started before 5.iv.21

See also

Other data checking functions: checkAllUnique(), getNNA(), getNOK()

Author

Chris Evans

Examples

if (FALSE) { ### this should map OK isOneToOne(1:5,letters[1:5]) ### 1:1 doesn't actually mean just one of each, ### just that the mapping is 1:1! isOneToOne(rep(1:5, 2), rep(letters[1:5], 2)) ### should throw an error as unequal length isOneToOne(1:26, letters[1:25]) ### but this is OK isOneToOne(1:26, c("1", letters[1:25])) ### but this is not as it's no longer 1:1 isOneToOne(c(1, 1:26), c("1", letters[1:25], "1")) ### same the other way around (essentially) isOneToOne(1:26,c("a",letters[1:25])) }