6
$\begingroup$

I have a list of list

A= {{0.620161, 0.320312, 0.94842, 1.11844, 1.12045, 1.12539, 1.13177, 
  1.13142, 1.15048, 1.23244, 0.721388, 0.708943, 0.750067, 0.744916, 
  0.720972, 0.674833, 1.29773, 1.29514}, {0.620161, 0.320312, 0.94842,
   1.11844, 1.12045, 1.12539, 1.13177, 1.13142, 1.15048, 0.721388, 
  0.750067, 0.744916, 0.720972, 0.674833, 1.29383, 
  1.29514}, {0.620161, 0.320312, 0.94842, 1.11844, 1.12045, 1.12539, 
  1.13177, 1.13142, 1.15048, 0.721388, 0.750067, 0.744916, 0.720972, 
  0.674833, 1.29383, 1.29514}, {0.620161, 0.320312, 0.94842, 1.11844, 
  1.12539, 1.13177, 1.13142, 1.15048, 0.721388, 0.750067, 0.744916, 
  0.720972, 0.674833, 1.29383, 1.29514}, .... }}.

The number of lists inside is not known. I want to replace all the elements of the first list by 20, all the elements of the second list by 20.1 and so on for all the lists.

I tried ReplaceList but with no success, can anyone help me with this?

Improve this question
$\endgroup$
2
  • $\begingroup$ You can start with list[[1,;;]]=20. It should be easy to generalize this in the way you need it. $\endgroup$
    – Grzegorz Rut
    Commented Oct 11, 2014 at 7:47
  • $\begingroup$ Yes, I know how to do list by list, but I don't know how to generalize or do it automatically for all the lists. Even the resources on Wolfram don't say most of the stuff that are discussed here. $\endgroup$
    – HuShu
    Commented Oct 11, 2014 at 8:10

6 Answers 6

Reset to default
6
$\begingroup$ 
ClearAll[miF, rPF, tBF];
miF = MapIndexed[20 + .1 (#2[[1]] - 1) &, #, {2}] &;

or

rPF = ReplacePart[#, {i_, _} :> (19.9 + 0.1 i)] &;

or

tbF = Block[{j = 0}, Table[ 0 i + 20 + 0.1 j++, {i, #}]] &;

Example:

A = {{0.620161, 0.320312, 0.94842, 1.11844, 1.12045, 1.12539, 1.13177,
     1.13142, 1.15048, 1.23244, 0.721388, 0.708943, 0.750067, 
    0.744916, 0.720972, 0.674833, 1.29773, 1.29514}, {0.620161, 
    0.320312, 0.94842, 1.11844, 1.12045, 1.12539, 1.13177, 1.13142, 
    1.15048, 0.721388, 0.750067, 0.744916, 0.720972, 0.674833, 
    1.29383, 1.29514}, {0.620161, 0.320312, 0.94842, 1.11844, 1.12045,
     1.12539, 1.13177, 1.13142, 1.15048, 0.721388, 0.750067, 0.744916,
     0.720972, 0.674833, 1.29383, 1.29514}, {0.620161, 0.320312, 
    0.94842, 1.11844, 1.12539, 1.13177, 1.13142, 1.15048, 0.721388, 
    0.750067, 0.744916, 0.720972, 0.674833, 1.29383, 1.29514}};

miF@A
(* {{20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20.},
    {20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1,  20.1, 20.1, 20.1, 20.1, 20.1,
     20.1, 20.1, 20.1, 20.1}, 
    {20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2,
     20.2, 20.2, 20.2, 20.2}, 
    {20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3,
     20.3, 20.3, 20.3}} *)

miF @ A == rPF @ A == tbF @ A
(* True  *)
Improve this answer
$\endgroup$
3
  • $\begingroup$ To a person who has never programmed before, this seems like magic. Thank you. $\endgroup$
    – HuShu
    Commented Oct 11, 2014 at 8:08
  • $\begingroup$ @Nilanjan, my pleasure. Welcome to Mathematica.SE. $\endgroup$
    – kglr
    Commented Oct 11, 2014 at 8:22
  • 2
    $\begingroup$ This is a rather inefficient method as you apply the function to every element rather than using a vector operation. $\endgroup$
    – Mr.Wizard
    Commented Oct 12, 2014 at 2:19
11
$\begingroup$

Another interesting approach:

i = 0;
Map[0 # + (20 + 0.1 i++) &, A]

{{20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 
  20., 20., 20., 20., 20.}, {20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1,
   20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1}, {20.2, 20.2,
   20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 
  20.2, 20.2, 20.2}, {20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 
  20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3}}
Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ The easiest approach, indeed $\endgroup$
    – Dr. belisarius
    Commented Oct 11, 2014 at 14:41
5
$\begingroup$

Using:

A = {{0.620161, 0.320312, 0.94842, 1.11844, 1.12045, 1.12539, 1.13177,
     1.13142, 1.15048, 1.23244, 0.721388, 0.708943, 0.750067, 
    0.744916, 0.720972, 0.674833, 1.29773, 1.29514}, {0.620161, 
    0.320312, 0.94842, 1.11844, 1.12045, 1.12539, 1.13177, 1.13142, 
    1.15048, 0.721388, 0.750067, 0.744916, 0.720972, 0.674833, 
    1.29383, 1.29514}, {0.620161, 0.320312, 0.94842, 1.11844, 1.12045,
     1.12539, 1.13177, 1.13142, 1.15048, 0.721388, 0.750067, 0.744916,
     0.720972, 0.674833, 1.29383, 1.29514}, {0.620161, 0.320312, 
    0.94842, 1.11844, 1.12539, 1.13177, 1.13142, 1.15048, 0.721388, 
    0.750067, 0.744916, 0.720972, 0.674833, 1.29383, 1.29514}};

A function to replace list:

rep[a_, v_] := MapThread[Table[#2, {Length@#1}] &, {a, v}]

so

rep[A, 19.9 + Range[Length[A]]/10]

yields:

{{20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 
  20., 20., 20., 20., 20.}, {20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1,
   20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1, 20.1}, {20.2, 20.2,
   20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 20.2, 
  20.2, 20.2, 20.2}, {20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 
  20.3, 20.3, 20.3, 20.3, 20.3, 20.3, 20.3}}
Improve this answer
$\endgroup$
1
  • $\begingroup$ ConstantArray would be better than Table here as the latter assumes a changing value. +1 nevertheless. $\endgroup$
    – Mr.Wizard
    Commented Oct 12, 2014 at 2:18
5
$\begingroup$

Borrowing from existing answers:

19.9 + Range@Length@A/10 + 0 A
Improve this answer
$\endgroup$
1
  • $\begingroup$ Why is it oftentimes so difficult to see the simple things? $\endgroup$
    – eldo
    Commented Oct 12, 2014 at 15:15
4
$\begingroup$ 
Table[A[[i, All]] = Table[20. + n/10, {n, 0, Length[A] - 1}][[i]], {i, 1, Length[A]}];

A // TableForm

enter image description here

Improve this answer
$\endgroup$
2
$\begingroup$

Two more:

19.9 + #1/10 + 0 #2 & @@@ Transpose@{Range@Length@A, A}

or

Transpose@{Range@Length@A, A} /. {x_, y_} :> 0 y + 19.9 + x/10
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.