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Let L be a list of several lists. And the length of each list is less than 1000.

You can create L by

L = Table[ConstantArray[1, RandomInteger[999]], 100000];

For example, let L looks like

{a list whose length is 340,  
 a list whose length is 525,  
 a list whose length is 829,  
 a list whose length is 422, 
......}

What I want to get is a list of integers

   M = {3, 5, 8, 4,......}

That is, if the length of an element is
between 0~99, assign 0,
between 100~199, assign 1,
...
between 900~999, assign 9.

In fact this can be done easily by Floor[Length[#]/100]&/@L

But I want to achieve it by built-in Replace.

Here is the code I did my best :

Replace[L, Table[_?(Length[#] < 100 k &) -> k - 1 /. k -> a, {a, 1, 10}], {1}] (*code of an  interest*)

But there is one thing I don't like about this code algorithmically.
This code computes Length[#] too often.
(Length is very fast, so just lucky this time.
What if the function is so slow.)
I want the Length to be computed only once for each element.

I tried codes like :

Replace[L, Module[{Leng}, Leng = Hold[Length[#]]; 
  Table[_?(Leng < 100 k &) -> T[k] /. k -> a, {a, 10}]], {1}] (*failed code*)

I tried Module,Block,With... all failed. Can you help me?
By modifying *code of an interest* slightly, can you show a code that computes Length only once for each element?

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    $\begingroup$ Is there a reason that you are so insistent on using Replace? Map seems like a natural thing to do here: Quotient[Length@#, 100] & /@ L. I would post that as the answer except that you seem to really want Replace. $\endgroup$
    – lericr
    May 13, 2022 at 13:27
  • $\begingroup$ Because.. If pattern1 matches then replace it to object1, If pattern2 matches then replace it to object2, If pattern3 match esthen replace it to object3... We can deal with this general problem. $\endgroup$
    – imida k
    May 14, 2022 at 4:02
  • $\begingroup$ Thank you, the example In my post was simple which means lucky enough to achieve the purpose with a short code. Here is my original intention and motivation : mathematica.stackexchange.com/questions/268227/… $\endgroup$
    – imida k
    May 14, 2022 at 4:07

1 Answer 1

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L = Table[ConstantArray[1, RandomInteger[999]], 100000];

m = Replace[L, {x_List :> Quotient[Length@x, 100]}, {1}];

$$\{7,2,8,8,6,5,1,5,2,1,1,3,6,7,2,4,7,2,1,8,2,3,8,6,5,2,0,7,4,9,6,3,2,4,6,5,7,7,9,9,\langle\langle 99920\rangle\rangle ,1,7,8,6,2,5,1,0,8,7,3,6,2,8,4,8,9,0,5,6,4,4,9,6,4,7,1,2,7,0,1,4,1,1,0,7,9,0,4,4\}$$

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