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Using Thread to map a two variable function

Nested Map and ApplyUsing MapIndexed only at certain elements of a listHow to use Map inside MapThread?Function output of two lists are only diagonal elementsUse map on a different levelHow to Map a function with HoldAll attributeHow to map a function on a variable listMapping a simple 2 variable function

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1

$begingroup$

I have the following function:

magicFunction[ lst_, val_ ] := Fold[ ((#2 < val) && #1) &, True, lst ]

It works

In[]=:magicFunction[6,7,9] 
Out[]= True

I want to Map the function. To do it I use Thread.

If I use Thread for a noname function with the list I want to map, the following is produced:

In[]:= Thread[f[6,7,2,3,1,2,9, 10,5]]
Out[]= f[6,7,9],f[2,3,10],f[1,2,5]

That is exactly what I want to do.

However, when I use magicFunction that is not what I get. If I use magicFunction I should get the following:

Out[]:= magicFunction[6,7,9],magicFunction[2,3,10],magicFunction[1,2,5]

which then evaluates to:

Out[]:= True, True, True

However what I get is the following:

In[]:= Thread[magicFunction[6,7,2,3,1,2,9, 10,5]]

Out[]:= 1,2<9,10,5&&2,3<9,10,5&&6,7<9,10,5

Does anybody know what I m doing wrong?

Thanks in advance!

map

share|improve this question

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

New contributor

Mavatanet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

add a comment | 

1

$begingroup$

I have the following function:

magicFunction[ lst_, val_ ] := Fold[ ((#2 < val) && #1) &, True, lst ]

It works

In[]=:magicFunction[6,7,9] 
Out[]= True

I want to Map the function. To do it I use Thread.

If I use Thread for a noname function with the list I want to map, the following is produced:

In[]:= Thread[f[6,7,2,3,1,2,9, 10,5]]
Out[]= f[6,7,9],f[2,3,10],f[1,2,5]

That is exactly what I want to do.

However, when I use magicFunction that is not what I get. If I use magicFunction I should get the following:

Out[]:= magicFunction[6,7,9],magicFunction[2,3,10],magicFunction[1,2,5]

which then evaluates to:

Out[]:= True, True, True

However what I get is the following:

In[]:= Thread[magicFunction[6,7,2,3,1,2,9, 10,5]]

Out[]:= 1,2<9,10,5&&2,3<9,10,5&&6,7<9,10,5

Does anybody know what I m doing wrong?

Thanks in advance!

map

share|improve this question

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

New contributor

Mavatanet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I have the following function:

magicFunction[ lst_, val_ ] := Fold[ ((#2 < val) && #1) &, True, lst ]

It works

In[]=:magicFunction[6,7,9] 
Out[]= True

I want to Map the function. To do it I use Thread.

If I use Thread for a noname function with the list I want to map, the following is produced:

In[]:= Thread[f[6,7,2,3,1,2,9, 10,5]]
Out[]= f[6,7,9],f[2,3,10],f[1,2,5]

That is exactly what I want to do.

However, when I use magicFunction that is not what I get. If I use magicFunction I should get the following:

Out[]:= magicFunction[6,7,9],magicFunction[2,3,10],magicFunction[1,2,5]

which then evaluates to:

Out[]:= True, True, True

However what I get is the following:

In[]:= Thread[magicFunction[6,7,2,3,1,2,9, 10,5]]

Out[]:= 1,2<9,10,5&&2,3<9,10,5&&6,7<9,10,5

Does anybody know what I m doing wrong?

Thanks in advance!

map

share|improve this question

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

New contributor

Mavatanet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

magicFunction[ lst_, val_ ] := Fold[ ((#2 < val) && #1) &, True, lst ]

In[]=:magicFunction[6,7,9] 
Out[]= True

In[]:= Thread[f[6,7,2,3,1,2,9, 10,5]]
Out[]= f[6,7,9],f[2,3,10],f[1,2,5]

Out[]:= magicFunction[6,7,9],magicFunction[2,3,10],magicFunction[1,2,5]

Out[]:= True, True, True

In[]:= Thread[magicFunction[6,7,2,3,1,2,9, 10,5]]

Out[]:= 1,2<9,10,5&&2,3<9,10,5&&6,7<9,10,5

share|improve this question

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

New contributor

Mavatanet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

New contributor

Mavatanet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

edited 7 hours ago

Carl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

New contributor

Mavatanet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

MavatanetMavatanet

5111 bronze badge

2 Answers
2

active

oldest

votes

3

$begingroup$

Your issue is that magicFunction evaluated before Thread could do anything. One idea to avoid this is to use Unevaluated:

Thread[Unevaluated @ magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

Another idea is to add an argument restriction to your magicFunction. Since Less only makes sense for numeric quantities, you can use:

Clear[magicFunction]
magicFunction[lst_, val_?NumericQ] := Fold[((#2<val)&&#1)&,True,lst]

Then:

Thread[magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

You can use MapThread:

MapThread[magicFunction, 6, 7, 2, 3, 1, 2, 9, 10, 5]

True, True, True

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

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$endgroup$

add a comment | 

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3

$begingroup$

Your issue is that magicFunction evaluated before Thread could do anything. One idea to avoid this is to use Unevaluated:

Thread[Unevaluated @ magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

Another idea is to add an argument restriction to your magicFunction. Since Less only makes sense for numeric quantities, you can use:

Clear[magicFunction]
magicFunction[lst_, val_?NumericQ] := Fold[((#2<val)&&#1)&,True,lst]

Then:

Thread[magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Your issue is that magicFunction evaluated before Thread could do anything. One idea to avoid this is to use Unevaluated:

Thread[Unevaluated @ magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

Another idea is to add an argument restriction to your magicFunction. Since Less only makes sense for numeric quantities, you can use:

Clear[magicFunction]
magicFunction[lst_, val_?NumericQ] := Fold[((#2<val)&&#1)&,True,lst]

Then:

Thread[magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

$endgroup$

add a comment | 

$begingroup$

Your issue is that magicFunction evaluated before Thread could do anything. One idea to avoid this is to use Unevaluated:

Thread[Unevaluated @ magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

Another idea is to add an argument restriction to your magicFunction. Since Less only makes sense for numeric quantities, you can use:

Clear[magicFunction]
magicFunction[lst_, val_?NumericQ] := Fold[((#2<val)&&#1)&,True,lst]

Then:

Thread[magicFunction[6,7,2,3,1,2,9,10,5]]

True, True, True

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

$endgroup$

Thread[Unevaluated @ magicFunction[6,7,2,3,1,2,9,10,5]]

Clear[magicFunction]
magicFunction[lst_, val_?NumericQ] := Fold[((#2<val)&&#1)&,True,lst]

Thread[magicFunction[6,7,2,3,1,2,9,10,5]]

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

answered 7 hours ago

Carl WollCarl Woll

89.6k33 gold badges117117 silver badges229229 bronze badges

3

$begingroup$

You can use MapThread:

MapThread[magicFunction, 6, 7, 2, 3, 1, 2, 9, 10, 5]

True, True, True

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges246246 silver badges492492 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

You can use MapThread:

MapThread[magicFunction, 6, 7, 2, 3, 1, 2, 9, 10, 5]

True, True, True

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges246246 silver badges492492 bronze badges

$endgroup$

add a comment | 

$begingroup$

You can use MapThread:

MapThread[magicFunction, 6, 7, 2, 3, 1, 2, 9, 10, 5]

True, True, True

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges246246 silver badges492492 bronze badges

$endgroup$

MapThread[magicFunction, 6, 7, 2, 3, 1, 2, 9, 10, 5]

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges246246 silver badges492492 bronze badges

answered 8 hours ago

kglrkglr

215k1010 gold badges246246 silver badges492492 bronze badges

answered 8 hours ago

kglrkglr

215k1010 gold badges246246 silver badges492492 bronze badges