Drop 1st element of {#a, #b, #c} &












4














I pass the following to a function:



{#a, #b, #c} &


These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:



{#b, #c} &


Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.



I looked at the full form and see that it's:



FullForm[{#a, #b, #c} &]





Function[List[Slot["a"], Slot["b"], Slot["c"]]]



I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.



Is there any way to get rid of the first element, #a, without rewriting the expression?






expression-manipulation slot







share|improve this question
























  • Instead of a blank you could replace Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.
    – Ruslan
    Nov 21 '18 at 6:51


















4














I pass the following to a function:



{#a, #b, #c} &


These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:



{#b, #c} &


Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.



I looked at the full form and see that it's:



FullForm[{#a, #b, #c} &]





Function[List[Slot["a"], Slot["b"], Slot["c"]]]



I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.



Is there any way to get rid of the first element, #a, without rewriting the expression?






expression-manipulation slot







share|improve this question
























  • Instead of a blank you could replace Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.
    – Ruslan
    Nov 21 '18 at 6:51
















4












4








4







I pass the following to a function:



{#a, #b, #c} &


These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:



{#b, #c} &


Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.



I looked at the full form and see that it's:



FullForm[{#a, #b, #c} &]





Function[List[Slot["a"], Slot["b"], Slot["c"]]]



I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.



Is there any way to get rid of the first element, #a, without rewriting the expression?






expression-manipulation slot







share|improve this question















I pass the following to a function:



{#a, #b, #c} &


These are keys that are used in a list of associations.
There are instances where I want to pass only the last 2 elements, such as:



{#b, #c} &


Right now, I just rewrite it. But it seems like there should be a way to do this by getting rid of the first slot.



I looked at the full form and see that it's:



FullForm[{#a, #b, #c} &]





Function[List[Slot["a"], Slot["b"], Slot["c"]]]



I thought I could somehow get inside and drop Slot["a"], but can't seem to do it. I tried replacing Slot["a"] with a blank, but I'm left with a comma at the beginning of the list. I also tried Apply to change Function to List and doing it in the result. I couldn't.



Is there any way to get rid of the first element, #a, without rewriting the expression?






expression-manipulation slot




expression-manipulation slot






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 '18 at 21:49









Kuba♦

103k12201516




103k12201516










asked Nov 20 '18 at 21:07









Mitchell Kaplan

1,6421126




1,6421126












  • Instead of a blank you could replace Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.
    – Ruslan
    Nov 21 '18 at 6:51




















  • Instead of a blank you could replace Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.
    – Ruslan
    Nov 21 '18 at 6:51


















Instead of a blank you could replace Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.
– Ruslan
Nov 21 '18 at 6:51






Instead of a blank you could replace Slot["a"]->Nothing. You'll get this Nothing in the resulting function object, but once the function is applied, Nothing will collapse.
– Ruslan
Nov 21 '18 at 6:51












2 Answers
2






active

oldest

votes


















7














You could pass it to Delete:



Delete[#, {1, 1}] &[{#a, #b, #c} &]



{#b, #c} &







share|improve this answer





























    5














    foo = {#a, #b, #c} &;
    

    
foo[[{1}, 2 ;;]]



    {#b, #c} &




    or, but only because we know there won't be any side effects: 



    Evaluate /@ Rest /@ foo



    {#b, #c} &




    or 



    foo /. {_Slot, rest__Slot} :> {rest}


    At the end consider using an operator form of a KeyDrop



    bar = KeyTake[{"a", "b", "c"}];
    

    
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>



    <|"b" -> 1, "c" -> 1|>







    share|improve this answer





















    • I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
      – Mitchell Kaplan
      Nov 20 '18 at 21:51










    • @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
      – Kuba♦
      Nov 20 '18 at 21:55












    • I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
      – Mitchell Kaplan
      Nov 21 '18 at 16:51











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    You could pass it to Delete:



    Delete[#, {1, 1}] &[{#a, #b, #c} &]



    {#b, #c} &







    share|improve this answer


























      7














      You could pass it to Delete:



      Delete[#, {1, 1}] &[{#a, #b, #c} &]



      {#b, #c} &







      share|improve this answer
























        7












        7








        7






        You could pass it to Delete:



        Delete[#, {1, 1}] &[{#a, #b, #c} &]



        {#b, #c} &







        share|improve this answer












        You could pass it to Delete:



        Delete[#, {1, 1}] &[{#a, #b, #c} &]



        {#b, #c} &








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 20 '18 at 21:09









        Coolwater

        14.7k32553




        14.7k32553























            5














            foo = {#a, #b, #c} &;
            

            
foo[[{1}, 2 ;;]]



            {#b, #c} &




            or, but only because we know there won't be any side effects: 



            Evaluate /@ Rest /@ foo



            {#b, #c} &




            or 



            foo /. {_Slot, rest__Slot} :> {rest}


            At the end consider using an operator form of a KeyDrop



            bar = KeyTake[{"a", "b", "c"}];
            

            
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>



            <|"b" -> 1, "c" -> 1|>







            share|improve this answer





















            • I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
              – Mitchell Kaplan
              Nov 20 '18 at 21:51










            • @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
              – Kuba♦
              Nov 20 '18 at 21:55












            • I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
              – Mitchell Kaplan
              Nov 21 '18 at 16:51
















            5














            foo = {#a, #b, #c} &;
            

            
foo[[{1}, 2 ;;]]



            {#b, #c} &




            or, but only because we know there won't be any side effects: 



            Evaluate /@ Rest /@ foo



            {#b, #c} &




            or 



            foo /. {_Slot, rest__Slot} :> {rest}


            At the end consider using an operator form of a KeyDrop



            bar = KeyTake[{"a", "b", "c"}];
            

            
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>



            <|"b" -> 1, "c" -> 1|>







            share|improve this answer





















            • I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
              – Mitchell Kaplan
              Nov 20 '18 at 21:51










            • @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
              – Kuba♦
              Nov 20 '18 at 21:55












            • I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
              – Mitchell Kaplan
              Nov 21 '18 at 16:51














            5












            5








            5






            foo = {#a, #b, #c} &;
            

            
foo[[{1}, 2 ;;]]



            {#b, #c} &




            or, but only because we know there won't be any side effects: 



            Evaluate /@ Rest /@ foo



            {#b, #c} &




            or 



            foo /. {_Slot, rest__Slot} :> {rest}


            At the end consider using an operator form of a KeyDrop



            bar = KeyTake[{"a", "b", "c"}];
            

            
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>



            <|"b" -> 1, "c" -> 1|>







            share|improve this answer












            foo = {#a, #b, #c} &;
            

            
foo[[{1}, 2 ;;]]



            {#b, #c} &




            or, but only because we know there won't be any side effects: 



            Evaluate /@ Rest /@ foo



            {#b, #c} &




            or 



            foo /. {_Slot, rest__Slot} :> {rest}


            At the end consider using an operator form of a KeyDrop



            bar = KeyTake[{"a", "b", "c"}];
            

            
(Rest /@ bar)@<|"a" -> 1, "b" -> 1, "c" -> 1|>



            <|"b" -> 1, "c" -> 1|>








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 20 '18 at 21:48









            Kuba♦

            103k12201516




            103k12201516












            • I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
              – Mitchell Kaplan
              Nov 20 '18 at 21:51










            • @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
              – Kuba♦
              Nov 20 '18 at 21:55












            • I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
              – Mitchell Kaplan
              Nov 21 '18 at 16:51


















            • I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
              – Mitchell Kaplan
              Nov 20 '18 at 21:51










            • @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
              – Kuba♦
              Nov 20 '18 at 21:55












            • I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
              – Mitchell Kaplan
              Nov 21 '18 at 16:51
















            I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
            – Mitchell Kaplan
            Nov 20 '18 at 21:51




            I accepted Coolwater's answer because it solved my problem simply. I appreciate the extra insight that your answer provided. Actually, I'll have to spend more time to really understand it though.
            – Mitchell Kaplan
            Nov 20 '18 at 21:51












            @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
            – Kuba♦
            Nov 20 '18 at 21:55






            @MitchellKaplan sure, btw, can you give a little insigth into how do you use it? Because it feels like a XY problem, maybe there is a better way anyway.
            – Kuba♦
            Nov 20 '18 at 21:55














            I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
            – Mitchell Kaplan
            Nov 21 '18 at 16:51




            I use a lot of lists of associations and very often I have to sum things by one or more keys. Why I asked this question: I work in reinsurance. My list has, among other things: treaty, effective date, line of business, IBNR. I'm estimating IBNR by all 3, but it has to balance to the total IBNR for effective/line. I need to sum both by all 3, but also by only effective/line. I call a function that does this, and I want to adjust the keys I'm using to eliminate treaty when I'm summing over just effective/line. Left some detail out to fit answer into comment.
            – Mitchell Kaplan
            Nov 21 '18 at 16:51


















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