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Taking the first element in a list of associations

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4

$begingroup$

I have the following list which is built up of associations, that is,

z = >, <

I am trying to to pick the first elements in each association such that it gives:

 >, <

to do that I do,

Map[First, z]

but this only returns the values, how can one fix this?

list-manipulation associations

share|improve this question

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

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add a comment | 

4

$begingroup$

I have the following list which is built up of associations, that is,

z = >, <

I am trying to to pick the first elements in each association such that it gives:

 >, <

to do that I do,

Map[First, z]

but this only returns the values, how can one fix this?

list-manipulation associations

share|improve this question

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

$endgroup$

add a comment | 

$begingroup$

I have the following list which is built up of associations, that is,

z = >, <

I am trying to to pick the first elements in each association such that it gives:

 >, <

to do that I do,

Map[First, z]

but this only returns the values, how can one fix this?

list-manipulation associations

share|improve this question

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

$endgroup$

z = >, <

 >, <

Map[First, z]

share|improve this question

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

share|improve this question

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

asked 11 hours ago

WilliamWilliam

1,10666 silver badges99 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:

z[[All, 1]]
(* "a", "b" -> 1 *)

(* without the list *)
z[[All, 1]]
(* 1, 4, 6, 10 *)

Alternatively you could use Take:

Map[Take[#, 1] &, z]
(* "a", "b" -> 1 *)

Other possibilites:

Map[#[[1]] &, z]
(* "a", "b" -> 1 *)

z[[All, ;; 1]]
(* "a", "b" -> 1 *)

In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)

share|improve this answer

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

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add a comment | 

2

$begingroup$

For completeness sake, a few more:

Extract[z, All, 1]
(* "a", "b" -> 1 *)

Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
(*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)

share|improve this answer

answered 6 hours ago

sakrasakra

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add a comment | 

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6

$begingroup$

You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:

z[[All, 1]]
(* "a", "b" -> 1 *)

(* without the list *)
z[[All, 1]]
(* 1, 4, 6, 10 *)

Alternatively you could use Take:

Map[Take[#, 1] &, z]
(* "a", "b" -> 1 *)

Other possibilites:

Map[#[[1]] &, z]
(* "a", "b" -> 1 *)

z[[All, ;; 1]]
(* "a", "b" -> 1 *)

In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)

share|improve this answer

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:

z[[All, 1]]
(* "a", "b" -> 1 *)

(* without the list *)
z[[All, 1]]
(* 1, 4, 6, 10 *)

Alternatively you could use Take:

Map[Take[#, 1] &, z]
(* "a", "b" -> 1 *)

Other possibilites:

Map[#[[1]] &, z]
(* "a", "b" -> 1 *)

z[[All, ;; 1]]
(* "a", "b" -> 1 *)

In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)

share|improve this answer

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

$endgroup$

add a comment | 

$begingroup$

You can use the fact that Part ([[…]]) preserves the keys when you supply a list of indices to extract:

z[[All, 1]]
(* "a", "b" -> 1 *)

(* without the list *)
z[[All, 1]]
(* 1, 4, 6, 10 *)

Alternatively you could use Take:

Map[Take[#, 1] &, z]
(* "a", "b" -> 1 *)

Other possibilites:

Map[#[[1]] &, z]
(* "a", "b" -> 1 *)

z[[All, ;; 1]]
(* "a", "b" -> 1 *)

In essence, anything that could in principle return multiple elements will preserve the keys (or, more generally, will preserve that level of the nested expression)

share|improve this answer

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

$endgroup$

z[[All, 1]]
(* "a", "b" -> 1 *)

(* without the list *)
z[[All, 1]]
(* 1, 4, 6, 10 *)

Map[Take[#, 1] &, z]
(* "a", "b" -> 1 *)

Map[#[[1]] &, z]
(* "a", "b" -> 1 *)

z[[All, ;; 1]]
(* "a", "b" -> 1 *)

share|improve this answer

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

answered 11 hours ago

Lukas LangLukas Lang

9,08411 gold badge1111 silver badges3434 bronze badges

2

$begingroup$

For completeness sake, a few more:

Extract[z, All, 1]
(* "a", "b" -> 1 *)

Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
(*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)

share|improve this answer

answered 6 hours ago

sakrasakra

3,2931414 silver badges3030 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

For completeness sake, a few more:

Extract[z, All, 1]
(* "a", "b" -> 1 *)

Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
(*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)

share|improve this answer

answered 6 hours ago

sakrasakra

3,2931414 silver badges3030 bronze badges

$endgroup$

add a comment | 

$begingroup$

For completeness sake, a few more:

Extract[z, All, 1]
(* "a", "b" -> 1 *)

Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
(*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)

share|improve this answer

answered 6 hours ago

sakrasakra

3,2931414 silver badges3030 bronze badges

$endgroup$

Extract[z, All, 1]
(* "a", "b" -> 1 *)

Cases[z, KeyValuePattern[x_ -> y_] :> x -> y]
(*"a", "b" -> 1, "g", "h" -> 4, "k", "l" -> 6, "s", "t" -> 10*)

share|improve this answer

answered 6 hours ago

sakrasakra

3,2931414 silver badges3030 bronze badges

answered 6 hours ago

sakrasakra

3,2931414 silver badges3030 bronze badges

answered 6 hours ago

sakrasakra

3,2931414 silver badges3030 bronze badges