Number of list elements less than a given integerCount number of sublists with a total not greater than a given maxHow do I select only those lists whose elements are all above a certain number?Defining functions of varying, iterative numbers of argumentsGeneral function taking general number of argumentsFunction taking arbitrary number of argumentsOverlap between lists of sublists with different dimensionsSpeed of a simple operation over a long listSpeed up Flatten[] of a large nested listDefining a Function programmatically
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Number of list elements less than a given integer
Count number of sublists with a total not greater than a given maxHow do I select only those lists whose elements are all above a certain number?Defining functions of varying, iterative numbers of argumentsGeneral function taking general number of argumentsFunction taking arbitrary number of argumentsOverlap between lists of sublists with different dimensionsSpeed of a simple operation over a long listSpeed up Flatten[] of a large nested listDefining a Function programmatically
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
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I'm trying to make a function whose input is a list, and that outputs a list for which the nth element is the number of elements in the input less than n.
So far, I have
l2 = Table[Length[Select[l1, # < n &]], n, 1, M]
Where l2 is the output list, l1 is the input list, and M is the desired length of l2.
This gives the correct output, but is there a faster way to do this, as this function can get rather slow for larger lists.
An example of the function:
2,2,3,3,7 —$>$ 0,0,2,4,4.
list-manipulation performance-tuning function-construction
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
$endgroup$
add a comment
|
$begingroup$
I'm trying to make a function whose input is a list, and that outputs a list for which the nth element is the number of elements in the input less than n.
So far, I have
l2 = Table[Length[Select[l1, # < n &]], n, 1, M]
Where l2 is the output list, l1 is the input list, and M is the desired length of l2.
This gives the correct output, but is there a faster way to do this, as this function can get rather slow for larger lists.
An example of the function:
2,2,3,3,7 —$>$ 0,0,2,4,4.
list-manipulation performance-tuning function-construction
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
$endgroup$
$begingroup$
Is your input list always sorted?
$endgroup$
– Carl Woll
7 hours ago
$begingroup$
Yes it is always sorted
$endgroup$
– volcanrb
7 hours ago
add a comment
|
$begingroup$
I'm trying to make a function whose input is a list, and that outputs a list for which the nth element is the number of elements in the input less than n.
So far, I have
l2 = Table[Length[Select[l1, # < n &]], n, 1, M]
Where l2 is the output list, l1 is the input list, and M is the desired length of l2.
This gives the correct output, but is there a faster way to do this, as this function can get rather slow for larger lists.
An example of the function:
2,2,3,3,7 —$>$ 0,0,2,4,4.
list-manipulation performance-tuning function-construction
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
$endgroup$
I'm trying to make a function whose input is a list, and that outputs a list for which the nth element is the number of elements in the input less than n.
So far, I have
l2 = Table[Length[Select[l1, # < n &]], n, 1, M]
Where l2 is the output list, l1 is the input list, and M is the desired length of l2.
This gives the correct output, but is there a faster way to do this, as this function can get rather slow for larger lists.
An example of the function:
2,2,3,3,7 —$>$ 0,0,2,4,4.
list-manipulation performance-tuning function-construction
list-manipulation performance-tuning function-construction
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
edited 8 hours ago
volcanrb
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
asked 8 hours ago
volcanrbvolcanrb
1954 bronze badges
1954 bronze badges
$begingroup$
Is your input list always sorted?
$endgroup$
– Carl Woll
7 hours ago
$begingroup$
Yes it is always sorted
$endgroup$
– volcanrb
7 hours ago
add a comment
|
$begingroup$
Is your input list always sorted?
$endgroup$
– Carl Woll
7 hours ago
$begingroup$
Yes it is always sorted
$endgroup$
– volcanrb
7 hours ago
$begingroup$
Is your input list always sorted?
$endgroup$
– Carl Woll
7 hours ago
$begingroup$
Is your input list always sorted?
$endgroup$
– Carl Woll
7 hours ago
$begingroup$
Yes it is always sorted
$endgroup$
– volcanrb
7 hours ago
$begingroup$
Yes it is always sorted
$endgroup$
– volcanrb
7 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
The BoolEval package can be used here to get a good combination of performance and clean syntax:
<< BoolEval`
Table[BoolCount[list < n], n, 5]
0, 0, 2, 4, 4
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
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$begingroup$
list = RandomReal[0, 100, 100000];
Accumulate[BinCounts[list, 0, Ceiling[Max[list]], 1]]
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
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add a comment
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Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The BoolEval package can be used here to get a good combination of performance and clean syntax:
<< BoolEval`
Table[BoolCount[list < n], n, 5]
0, 0, 2, 4, 4
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
$endgroup$
add a comment
|
$begingroup$
The BoolEval package can be used here to get a good combination of performance and clean syntax:
<< BoolEval`
Table[BoolCount[list < n], n, 5]
0, 0, 2, 4, 4
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
$endgroup$
add a comment
|
$begingroup$
The BoolEval package can be used here to get a good combination of performance and clean syntax:
<< BoolEval`
Table[BoolCount[list < n], n, 5]
0, 0, 2, 4, 4
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
$endgroup$
The BoolEval package can be used here to get a good combination of performance and clean syntax:
<< BoolEval`
Table[BoolCount[list < n], n, 5]
0, 0, 2, 4, 4
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
answered 7 hours ago
C. E.C. E.
55.6k3 gold badges108 silver badges221 bronze badges
55.6k3 gold badges108 silver badges221 bronze badges
add a comment
|
add a comment
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$begingroup$
list = RandomReal[0, 100, 100000];
Accumulate[BinCounts[list, 0, Ceiling[Max[list]], 1]]
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
$endgroup$
add a comment
|
$begingroup$
list = RandomReal[0, 100, 100000];
Accumulate[BinCounts[list, 0, Ceiling[Max[list]], 1]]
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
$endgroup$
add a comment
|
$begingroup$
list = RandomReal[0, 100, 100000];
Accumulate[BinCounts[list, 0, Ceiling[Max[list]], 1]]
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
$endgroup$
list = RandomReal[0, 100, 100000];
Accumulate[BinCounts[list, 0, Ceiling[Max[list]], 1]]
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
answered 6 hours ago
Henrik SchumacherHenrik Schumacher
68.5k5 gold badges98 silver badges191 bronze badges
68.5k5 gold badges98 silver badges191 bronze badges
add a comment
|
add a comment
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$begingroup$
Is your input list always sorted?
$endgroup$
– Carl Woll
7 hours ago
$begingroup$
Yes it is always sorted
$endgroup$
– volcanrb
7 hours ago