A list of number modulus their indices in the list

Question

A simple one: Take a list of positive integers as input and output the numbers modulus their 1-based index in the list.

If the input integers are {a, b, c, d, e, f, g} then the output should be {a%1, b%2, c%3, d%4, e%5, f%6, g%7} where % is the modulus operator.

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Test cases:

10  9  8  7  6  5  4  3  2  1
 0  1  2  3  1  5  4  3  2  1

8 18  6 11 14  3 15 10  6 19 12  3  7  5  5 19 12 12 14  5
0  0  0  3  4  3  1  2  6  9  1  3  7  5  5  3 12 12 14  5

1
0

1  1
0  1

Answer 1

Haskell, 20 bytes

($[1..]).zipWith mod

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A trick to flip I learned from a golf of Anders Kaseorg.

Answer 2

Operation Flashpoint scripting language, 73 bytes

f={l=_this;r=[];i=0;while{i<count l}do{r=r+[(l select i)%(i+1)];i=i+1};r}

Call with:

numList = [10, 9, 8, 7, 6, 5, 4, 3, 2, 1];
hint format["%1\n%2", numList, numList call f];

Output:

Answer 3

Python 2, 35 bytes

i=1
for x in input():print x%i;i+=1

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Counts the index up manually, as per a tip of mine.

Answer 4

Jelly, 2 bytes

%J

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Explanation:

%J
 J List 1 .. len(input). This is results in a list of the indexes.
%  Modulo.

Basically, the code modulos the original list by the list of indexes.

Answer 5

R, 24 18 bytes

pryr::f(x%%seq(x))

Evaluates to the function:

function (x) 
x%%seq(x)

Which uses seq_along() to create a vector of the same length as x, starting at 1, and then %% to take the modulo.

Default behaviour of seq when presented with a vector is seq(along.with = x) which is the same output as seq_along(x), but 6 bytes shorter.

Answer 6

R, 27 bytes

x=scan();cat(x%%1:sum(1|x))

saved 5 bytes thanks to @Jarko

saved 4 more thanks to @Giuseppe

saved 2 more thanks to @Taylor Scott

Saved 2 more thanks to @returnbull

Answer 7

MATL, 4, 3 bytes

tf\

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One byte saved thanks to @LuisMendo!

Answer 8

APL (Dyalog), 5 bytes

⍳∘≢|⊢

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⍳ the indices

∘ of

≢ the length of the argument

| that modulus

⊢ the argument

Answer 9

Cubix, 19 bytes

;ww.1I!@s%Ow;)Sow.$

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    ; w
    w .
1 I ! @ s % O w
; ) S o w . $ .
    . .
    . .

Watch It Run

A fairly straight forward implementation.

• 1 push 1 to the stack to start the index
• I!@ get the integer input and halt if 0
• s%Ow swap the index up, mod, output result and change lane
• ;) remove result and increment index
• Sow push 32, output space and change lane (heading down from o)
• $O jump the output
• w;w change lange, remove 32 from stack and change lane onto the I input

Answer 10

05AB1E, 2 bytes

ā%

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ā  # Push the range(1, len(a) + 1)
 % # Mod each element in the input by the same one in this list

Answer 11

Mathematica, 22 bytes

#&@@@Mod~MapIndexed~#&

One more Mathematica approach.

Answer 12

Starry, 75 70 bytes

      +`  , + +   *    +  + +      +*   +    *  .               + + .'

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Explanation

This is an infinite loop that keeps reading numbers from the input and increasing a counter initiallized at 1. For each pair of input and counter, the modulus is computed and printed.

To end the loop when input has been exhausted, the following trick is used. When no more input is available, trying to read one more number gives a 0. Thus, we divide the read number by itself, and if it is 0 the program ends with an error. Else we discard the result and continue.

      +              Push 1. This is the initial value of the counter
`                    Mark label
  ,                  Read number from input and push it. Gives 0 if no more input
 +                   Duplicate top of the stack
 +                   Duplicate top of the stack
   *                 Pop two numbers and push their division. Error if divisor is 0
    +                Pop (discard) top of the stack
  +                  Swap top two numbers
 +                   Duplicate top of the stack
      +              Push 1
*                    Pop two numbers and push their sum. This increases the counter
   +                 Rotate stack down, to move increased counter to bottom
    *                Pop two numbers and push their modulus
  .                  Pop a number and print it as a number
               +     Push 10
 +                   Duplicate top of the stack
 .                   Pop a number (10) and print it as ASCII character (newline)
'                    If top of the stack is non-zero (it is, namely 10) go to label

Answer 13

Brachylog, 9 bytes

l⟦₁;?↔z%ᵐ

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Explanation

l⟦₁          [1, ..., length(Input)]
   ;?↔z      Zip the Input with that range
       %ᵐ    Map mod

Answer 14

R, 22 bytes

pryr::f(x%%1:sum(x|1))

R performs 1:length(x) before doing the modulus.

Answer 15

Itr, 5 bytes

#äL¹%

online interpreter

Explanation

#     ; read the list from standard input
 äL   ; push its length
   ¹  ; converted to 1-based range
    % ; point-wise modulo
      ; implicitly output list

---

Itr, 4 bytes

#L¹%

The ä before the L is no-longer necessary in newer versions

Answer 16

Japt -m, 5 4 3 bytes

uVÄ

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-1 byte thanks to ETHproductions

Answer 17

Python 2, 42 bytes

lambda l:[v%(i+1) for i,v in enumerate(l)]

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Answer 18

Haskell, 22 bytes

zipWith(flip mod)[1..]

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Also: flip(zipWith mod)[1..].

Answer 19

Mathematica, 21 bytes

#~Mod~Range@Length@#&

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or 20 bytes (by Martin)

#~Mod~Range@Tr[1^#]&

Answer 20

Excel VBA, 59 46 Bytes

Golfed

Anonymous VBE Immediate window funtion that takes a space ( ) delimited array string as input from range [A1] and output the numbers modulus their 1-based index in the starting list to the VBE immediate window

For Each n In Split([A1]):i=i+1:?n Mod i;:Next

Input / Output:

[A1]="10 9 8 7 6 5 4 3 2 1" ''# or manually set the value
For Each n In Split([A1]):i=i+1:?n Mod i;:Next
 0  1  2  3  1  5  4  3  2  1 

Old Subroutine version

Subroutine that takes input as a passed array and outouts to the VBE immediate window.

Sub m(n)
For Each a In n
i=i+1
Debug.?a Mod i;
Next
End Sub

Input / Ouput:

m Array(10,9,8,7,6,5,4,3,2,1)
 0  1  2  3  1  5  4  3  2  1 

Ungolfed

Option Private Module
Option Compare Binary
Option Explicit
Option Base 0 ''# apparently Option Base 1 does not work with ParamArrays

Public Sub modIndex(ParamArray n() As Variant)
    Dim index As Integer
    For index = LBound(n) To UBound(n)
        Debug.Print n(index) Mod (index + 1);
    Next index
End Sub

Input / Output:

Call modIndex(10,9,8,7,6,5,4,3,2,1)
 0  1  2  3  1  5  4  3  2  1 

Answer 21

Thunno 2, 2 bytes

ż%

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Explanation

ż%  # Implicit input
 $  # Vectorised modulo
ż   # List [1..length]
    # Implicit output

Answer 22

Rust, 46 bytes

|v|v.iter().zip(1..).map(|(n,i)|n%i).collect()

Rust Playground

Answer 23

CJam, 9 bytes

{_,,:).%}

Anonymous block that expects an array on the stack and replaces it by the output array.

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Explanation

{       }    e# Define block
 _           e# Duplicate
  ,          e# Length
   ,         e# Range, 0-based
    :)       e# Add 1 to each entry
      .%     e# Vectorized modulus

Answer 24

J, 9 bytes

>:@i.@#|[

1 ... n | original list

| is mod

Answer 25

JavaScript (ES6), 22 bytes

a=>a.map((x,y)=>x%++y)

Answer 26

AWK, 13

{print $1%NR}

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Answer 27

tcl, 35

lmap l $L {puts [expr $l%[incr i]]}

demo

Answer 28

GNU APL 1.2, 9 bytes

(⍳⍴R)|R←⎕

APL operates from right to left, hence the parentheses.

R←⎕ assigns user input to vector R.

⍴R gives the length of the vector; ⍳⍴R gives a vector with all numbers from 1 to that length (so the indices).

| is the mod operator (a|b yields b%a). APL operates on arrays, so the code snippet a vector containing each element from the user's input mod its index.

Answer 29

Pari/GP, 22 bytes

a->[a[n]%n|n<-[1..#a]]

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Answer 30

Pyth, 5

.e%bh

Online test.

    hk     # 1-based index of (implicit) lambda variable
   b       # element
  %        # element mod (1-based index)
.e    Q    # enumerated map over (implicit) input