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Position of a subset by its length and member?

Eduardo Mendes
Posted 10 years ago

Hello

I am struggling to understand how Position and Select work when using patterns. Although I've got many examples right (thanks to the help of the members of this list) I still get stuck trying to find the right pattern (rule?) for some examples.

list={{y3}, {y0 y3, y3}, {y0 y1 y3, y1 y3}, {}}

I need to find the position of {y3} so I tried

Position[list, _?((Length[#] == 1  &) \[And] (MemberQ[#, y3] &))]

but it didn't work. I am definitely missing something since ((Length[lista[[1]]] == 1) \[And] ( MemberQ[lista[[1]], y3])) returns true as expected. Please tell what I am missing.

Many thanks

Cheers

Ed
POSTED BY: Eduardo Mendes
8 Replies
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Eduardo Mendes
Eduardo Mendes
Posted 9 years ago

Many thanks. I can see the difference.
POSTED BY: Eduardo Mendes

...and have a look at

Hold[Position[ list, _?(Length[#] == 1 \[And] MemberQ[#, y3] &)]] // FullForm

and

Hold[Position[ list, _?(Length[#] == 1 \[And] MemberQ[#, y3]) &]] // FullForm

note the different expressions for Function[...]
POSTED BY: Hans Dolhaine
Eduardo Mendes
Eduardo Mendes
Posted 10 years ago

I see. Many thanks.

Cheers

Ed
POSTED BY: Eduardo Mendes

I suppose it is important whether the & is within the parenthesis or not:

Position[list, _?(Length[#] == 1 \[And] MemberQ[#, y3] &)]

and 

Position[list, _?(Length[#] == 1 \[And] MemberQ[#, y3]) &]

does not work.

Cheers,

M.
POSTED BY: Marco Thiel
Eduardo Mendes
Eduardo Mendes
Posted 10 years ago

Hi Marco

Many thanks. So the problem was the extra parentheses? (() and ()) is not allowed, is it? Do you know why not?

Cheers

Ed
POSTED BY: Eduardo Mendes
Eduardo Mendes
Eduardo Mendes
Posted 10 years ago

Hi Marco

Many thanks again. Would you mind telling me why MemberQ[list[[1]] and not in the way I did?

Yes, I want to generalize by using multiple conditions.

Cheers

Ed
POSTED BY: Eduardo Mendes

Hi Ed,

you are actually right. I copied the wrong line:

Position[list, _?(Length[#] == 1 \[And] MemberQ[#, y3] &)]

is better. The only problem was a matter of parenthesis.

Sorry,

Marco
POSTED BY: Marco Thiel

Hi Ed,

you were actually quite close:

Position[list, _?(Length[#] == 1 \[And] MemberQ[list[[1]], y3] &)]

Of course, there are easier ways in this case:

Position[list, {y3}]

but you might be asking because you wan too generalise this.

Cheers,

Marco
POSTED BY: Marco Thiel
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