isOneToOne: Data checking function to check if the contents of two...

Description Usage Arguments Value Background Acknowledgements History/development log Author(s) See Also Examples

View source: R/isOneToOne.R

Description

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

Usage

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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:

x y
1 Male
2 Female
3 Other
2 Female
2 Female
1 Male
3 Other
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.
x y
1 Male
2 Feemale
3 Other
2 Female
2 Female
1 male
3 Other
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

Author(s)

Chris Evans

See Also

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

Examples

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## Not run: 
### 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]))

## End(Not run)

cpsyctc/CECPfuns documentation built on April 15, 2021, 3:26 a.m.