Odtnhj

Say I have a sorted list of integers

RandomInteger[1, 100000, 10000] // Sort // Short

I want to construct another list whose $m$-th element is the number of elements in the original list that are less than or equal to $m$:

Table[Length@Select[%, LessEqualThan[m]], m, 10000]

This is terribly inefficient, but for some reason I cannot come up with a better a approach. What's a better way to accomplish this? This seems to be a fairly standard exercise, so there should be plenty of duplicates, but I can find none.
I am probably missing a key word...

You can use the usual UnitStep + Total tricks:

r1 = Table[Total[UnitStep[m-s]], m,10000]; //AbsoluteTiming

r2 = Table[Length@Select[s,LessEqualThan[m]],m,10000];//AbsoluteTiming

r1 === r2

0.435358, Null

41.4357, Null

True

Update

As @J42161217 points out, you can take advantage of the fact that the data is sorted to speed things up. He used Differences. Here is a version that uses Nearest instead:

mincounts[s_] := With[
 
 unique = DeleteDuplicates@Nearest[s->"Element",s][[All,-1]],
 counts = Prepend[0] @ DeleteDuplicates@Nearest[s->"Index",s][[All,-1]]
 ,
 With[near = Nearest[unique->"Index", Range @ Length @ s][[All,1]],
 counts[[1+near-UnitStep[unique[[near]]-Range@Length@s-1]]]
 ]
]

Comparison:

SeedRandom[1];
s=RandomInteger[1,100000,10000]//Sort;

(* my first answer *)
r1 = Table[Total[UnitStep[m-s]], m,10000]; //AbsoluteTiming
(* J42161217's answer *)
r2 = Flatten[
 Join[
 Table[0, s[[1]] - 1], 
 Table[Table[i, Differences[s][[i]]], i, Length[Select[s, # <= 10000 &]]]
 ]
][[;;10000]]; // AbsoluteTiming
(* using Nearest *)
r3 = mincounts[s]; //AbsoluteTiming

r1 === r2 === r3

0.432897, Null

0.122198, Null

0.025923, Null

True

BinCounts and Accumulate combination is faster than all the methods posted so far:

r4 = Accumulate @ BinCounts[s, 1, 1 + 10000, 1]; // RepeatedTiming // First 

0.00069

versus Henrik Schumacher's mySparseArray, Carl Woll's mincounts and J42161217's Differences-based method:

r5 = Accumulate[mySparseArray[Partition[s, 1] -> 1, s[[-1]], Total, 0][[
 1 ;; Length[s]]]
 ]; // RepeatedTiming // First

 0.00081

r3 = mincounts[s]; // RepeatedTiming // First

0.016

r2 = Flatten[Join[Table[0, s[[1]] - 1], 
 Table[Table[i, Differences[s][[i]]], i, 
     Length[Select[s, # <= 10000 &]]]]][[;; 10000]]; // 
 RepeatedTiming // First 

0.149

r2 == r3 == r4 == r5

True

I think this is at least x3 faster than Mr. Carl Woll's answer

Can anybody compare my timing?

r3 = Flatten[Join[Table[0, s[[1]] - 1], 
Table[Table[i, Differences[s][[i]]], i, 
 Length[Select[s, # <= 10000 &]]]]][[;;10000]]; // AbsoluteTiming 

0.157123, Null

Using MapThread the same code is way faster

r6 = Flatten[
 Join[Table[0, s[[1]] - 1], 
 MapThread[
 Table, Range[t = Length[Select[s, # <= 10000 &]]], 
 Differences[s][[1 ;; t]]]]][[;; 10000]]; // AbsoluteTiming 

r6===r3 

0.008387, Null

True

s = Sort[RandomInteger[1, 100000, 10000]];

Let us just imagine for the moment that the target list r is supposed to have length 100000 (we can truncate it afterwards). Then for each entry i in the list s, the list r needs to have a jump of height 1 at position i. The jumps are the "derivative" of r (in a discrete sense) and the antiderivative can be obtained with Accumulate. In order to get the list of jumps, we need additive matrix assembly.

This can be done with this function:

mySparseArray[rules_, dims_, f_: Total, background_: 0.] := 
 If[(Head[rules] === Rule) && (rules[[1]] === ),
 rules[[2]],
 With[spopt = SystemOptions["SparseArrayOptions"],
 Internal`WithLocalSettings[
 SetSystemOptions[
 "SparseArrayOptions" -> "TreatRepeatedEntries" -> f],
 SparseArray[rules, dims, background],
 SetSystemOptions[spopt]]
 ]
 ]

So, in total, r can be obtained as follows:

r4 = Accumulate[
 mySparseArray[Partition[s, 1] -> 1, s[[-1]], Total, 0][[1 ;; Length[s]]]
 ]; // RepeatedTiming // First

0.00055

For comparison:

r3 = Flatten[
 Join[Table[0, s[[1]] - 1], 
 Table[Table[i, Differences[s][[i]]], i, 
 Length[Select[s, # <= 10000 &]]]]][[;; 10000]]; // 
 RepeatedTiming // First
r3 == r4

0.116

True