6
$\begingroup$

I have the following list containing 2-element sublists:

list={{0.01, 0.037348}, {0.03, 0.165434}, {0.1, 0.263921}, {0.3, 
  0.560191}, {1., 0.968857}, {3., 1.50965}, {10., 2.36502}, {30., 
  3.07659}, {100., 3.73412}, {300., 4.4931}, {1000., 5.06818}, {3000.,
   5.65423}, {10000., 6.00944}}

How can I get the natural logarithm (Log[]) of the first element of each sublist (e.g. 0.01, 0.03.....10000) only while leaving the second one intact?. I would like this operation to apply to other sublists of different lengths.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ {Log[First[#]],Last[#]}&/@list will make a new list with log of the first element. $\endgroup$
    – flinty
    Jun 28, 2020 at 1:26
  • $\begingroup$ This works great! @flinty. Thank you very much !!! $\endgroup$
    – John
    Jun 28, 2020 at 1:29
  • 2
    $\begingroup$ Or {Log@#1, #2} & @@@ list. There are many similar questions and this will be signed as a duplicate. $\endgroup$
    – Artes
    Jun 28, 2020 at 1:50
  • 1
    $\begingroup$ list[[All,1]]=Log@list[[All,1]] $\endgroup$
    – OkkesDulgerci
    Jun 28, 2020 at 2:33
  • $\begingroup$ If you're ultimately planning to plot this data, you may want to look at the documentation for ListLogPlot, ListLogLinearPlot, and/or ListLogLogPlot. $\endgroup$
    – Michael Seifert
    Jun 28, 2020 at 14:04

6 Answers 6

Reset to default
6
$\begingroup$

This should be fast for large lists:

Transpose[{Log @ #, #2} & @@ Transpose[#]] & @ list

{{-4.60517, 0.037348}, {-3.50656, 0.165434}, {-2.30259, 0.263921}, {-1.20397, 0.560191}, 
{0., 0.968857}, {1.09861, 1.50965}, {2.30259, 2.36502}, {3.4012, 3.07659}, {4.60517,  3.73412},
{5.70378, 4.4931}, {6.90776, 5.06818}, {8.00637, 5.65423}, {9.21034, 6.00944}}
Improve this answer
$\endgroup$
10
$\begingroup$

You may use MapAt.

MapAt[Log, {All, 1}]@list

{{-4.60517, 0.037348}, {-3.50656, 0.165434}, 
 {-2.30259, 0.263921}, {-1.20397, 0.560191}, 
 {0., 0.968857}, {1.09861, 1.50965}, 
 {2.30259, 2.36502}, {3.4012, 3.07659}, 
 {4.60517, 3.73412}, {5.70378, 4.4931}, 
 {6.90776, 5.06818}, {8.00637, 5.65423}, 
 {9.21034, 6.00944}}

Hope this helps.

Improve this answer
$\endgroup$
7
$\begingroup$

In addition, in v 12.0 and greater, you may use SubsetMap

SubsetMap[Log, list, {All,1}]

{{-4.60517, 0.037348}, {-3.50656, 0.165434}, {-2.30259, 0.263921}, {-1.20397, 0.560191}, {0., 0.968857}, {1.09861, 1.50965}, {2.30259, 2.36502}, {3.4012, 3.07659}, {4.60517, 3.73412}, {5.70378, 4.4931}, {6.90776, 5.06818}, {8.00637, 5.65423}, {9.21034, 6.00944}}

Improve this answer
$\endgroup$
3
$\begingroup$ 
newlist=Table[{Log[list[[k]]],list[[k]]},{k,1,Length[list]}]

is inelegant but will work.

Improve this answer
$\endgroup$
2
$\begingroup$ 
Query[All, {1 -> Log}] @ list

{{-4.60517, 0.037348}, {-3.50656, 0.165434}, {-2.30259, 0.263921}, {-1.20397, 0.560191}, {0., 0.968857}, {1.09861, 1.50965}, {2.30259, 2.36502}, {3.4012, 3.07659}, {4.60517, 3.73412}, {5.70378, 4.4931}, {6.90776, 5.06818}, {8.00637, 5.65423}, {9.21034, 6.00944}}

Map Log over first and Log10 over last column:

Query[All, {1 -> Log, -1 -> Log10}] @ list;

To Query rows omit the All:

Query[{1 -> Log, -1 -> Log10}] @ list
Improve this answer
$\endgroup$
1
$\begingroup$ 
list={{0.01, 0.037348}, {0.03, 0.165434}, {0.1, 0.263921}, {0.3, 
  0.560191}, {1., 0.968857}, {3., 1.50965}, {10., 2.36502}, {30., 
  3.07659}, {100., 3.73412}, {300., 4.4931}, {1000., 5.06818}, {3000.,
   5.65423}, {10000., 6.00944}}

Cases[list, {a_, b_} :> {Log[a], b}, 1]
Replace[list, {a_, b_} :> {Log[a], b}, {1}]
SequenceReplace[list, {{a_, b_}} :> {Log[a], b}]
{list[[All, 1]] // Log, list[[All, 2]]} // Transpose
MapThread[Compose, {{Log, Identity}, #}] & /@ list

{{-4.60517, 0.037348}, {-3.50656, 0.165434}, {-2.30259, 0.263921}, {-1.20397, 0.560191}, {0., 0.968857}, {1.09861, 1.50965}, {2.30259, 2.36502}, {3.4012, 3.07659}, {4.60517, 3.73412}, {5.70378, 4.4931}, {6.90776, 5.06818}, {8.00637, 5.65423}, {9.21034, 6.00944}}

Improve this answer
$\endgroup$

Your Answer

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

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