Counting certain elements in listsSplitting up delimited data in listsCounting function, comparing listsTake the next element in a nested listIssue with very large lists in Mathematica“renormalising” a listTiming and memory use is critical:fast partitioning of binary sparse arrayReplicate sublist in new listLess than Nothingrebinning of dataCounting elements in a list
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Counting certain elements in lists
Splitting up delimited data in listsCounting function, comparing listsTake the next element in a nested listIssue with very large lists in Mathematica“renormalising” a listTiming and memory use is critical:fast partitioning of binary sparse arrayReplicate sublist in new listLess than Nothingrebinning of dataCounting elements in a list
$begingroup$
I have the following data set:
n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];
Dimensions@data1
1000, 2
I want to count how often data1[[All, 2]] == 1.
My solution, which seems to be wrong:
result1 = Table[
Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
i, 1, n1
];
Dimensions@result1
1000
Now I have 50 appended lists of data1 (= data2).
n3 = 50;
data2 = Array[0 &, n3];
Do[
data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
, i, 1, n3
];
Dimensions@data2
50, 1000, 2
I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.
My solution for this more complicated list:
result2 = Array[0 &, n3];
Do[
result2[[j]] =
Table[
Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
i, 1, n1
];
, j, 1, n3
];
Dimensions@result2
50, 1000
How can I replace the Do loops in both cases and improve the performance?
list-manipulation
asked 1 hour ago
liolio
1,135217
$endgroup$
add a comment |
$begingroup$
I have the following data set:
n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];
Dimensions@data1
1000, 2
I want to count how often data1[[All, 2]] == 1.
My solution, which seems to be wrong:
result1 = Table[
Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
i, 1, n1
];
Dimensions@result1
1000
Now I have 50 appended lists of data1 (= data2).
n3 = 50;
data2 = Array[0 &, n3];
Do[
data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
, i, 1, n3
];
Dimensions@data2
50, 1000, 2
I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.
My solution for this more complicated list:
result2 = Array[0 &, n3];
Do[
result2[[j]] =
Table[
Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
i, 1, n1
];
, j, 1, n3
];
Dimensions@result2
50, 1000
How can I replace the Do loops in both cases and improve the performance?
list-manipulation
asked 1 hour ago
liolio
1,135217
$endgroup$
add a comment |
$begingroup$
I have the following data set:
n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];
Dimensions@data1
1000, 2
I want to count how often data1[[All, 2]] == 1.
My solution, which seems to be wrong:
result1 = Table[
Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
i, 1, n1
];
Dimensions@result1
1000
Now I have 50 appended lists of data1 (= data2).
n3 = 50;
data2 = Array[0 &, n3];
Do[
data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
, i, 1, n3
];
Dimensions@data2
50, 1000, 2
I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.
My solution for this more complicated list:
result2 = Array[0 &, n3];
Do[
result2[[j]] =
Table[
Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
i, 1, n1
];
, j, 1, n3
];
Dimensions@result2
50, 1000
How can I replace the Do loops in both cases and improve the performance?
list-manipulation
asked 1 hour ago
liolio
1,135217
$endgroup$
I have the following data set:
n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];
Dimensions@data1
1000, 2
I want to count how often data1[[All, 2]] == 1.
My solution, which seems to be wrong:
result1 = Table[
Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
i, 1, n1
];
Dimensions@result1
1000
Now I have 50 appended lists of data1 (= data2).
n3 = 50;
data2 = Array[0 &, n3];
Do[
data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
, i, 1, n3
];
Dimensions@data2
50, 1000, 2
I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.
My solution for this more complicated list:
result2 = Array[0 &, n3];
Do[
result2[[j]] =
Table[
Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
i, 1, n1
];
, j, 1, n3
];
Dimensions@result2
50, 1000
How can I replace the Do loops in both cases and improve the performance?
list-manipulation
list-manipulation
asked 1 hour ago
liolio
1,135217
asked 1 hour ago
liolio
1,135217
edited 1 hour ago
lio
asked 1 hour ago
liolio
1,135217
asked 1 hour ago
liolio
1,135217
asked 1 hour ago
liolio
1,135217
1,135217
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], 2];
answered 1 hour ago
MarcoBMarcoB
37.5k556113
$endgroup$
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
1 hour ago
add a comment |
$begingroup$
r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2]
Short @ %
5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5
r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions
50, 1000
answered 1 hour ago
kglrkglr
188k10205422
$endgroup$
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], 2];
answered 1 hour ago
MarcoBMarcoB
37.5k556113
$endgroup$
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
1 hour ago
add a comment |
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], 2];
answered 1 hour ago
MarcoBMarcoB
37.5k556113
$endgroup$
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
1 hour ago
add a comment |
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], 2];
answered 1 hour ago
MarcoBMarcoB
37.5k556113
$endgroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], 2];
answered 1 hour ago
MarcoBMarcoB
37.5k556113
edited 1 hour ago
answered 1 hour ago
MarcoBMarcoB
37.5k556113
answered 1 hour ago
MarcoBMarcoB
37.5k556113
answered 1 hour ago
MarcoBMarcoB
37.5k556113
37.5k556113
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
1 hour ago
add a comment |
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
1 hour ago
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
This is great ...
$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea for
result2 improvement?$endgroup$
– lio
1 hour ago
$begingroup$
Do you have an idea for
result2 improvement?$endgroup$
– lio
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as your
result1 and result2 respectively.$endgroup$
– MarcoB
1 hour ago
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as your
result1 and result2 respectively.$endgroup$
– MarcoB
1 hour ago
add a comment |
$begingroup$
r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2]
Short @ %
5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5
r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions
50, 1000
answered 1 hour ago
kglrkglr
188k10205422
$endgroup$
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
add a comment |
$begingroup$
r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2]
Short @ %
5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5
r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions
50, 1000
answered 1 hour ago
kglrkglr
188k10205422
$endgroup$
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
add a comment |
$begingroup$
r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2]
Short @ %
5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5
r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions
50, 1000
answered 1 hour ago
kglrkglr
188k10205422
$endgroup$
r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2]
Short @ %
5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5
r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions
50, 1000
answered 1 hour ago
kglrkglr
188k10205422
edited 1 hour ago
answered 1 hour ago
kglrkglr
188k10205422
answered 1 hour ago
kglrkglr
188k10205422
answered 1 hour ago
kglrkglr
188k10205422
188k10205422
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
add a comment |
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
$begingroup$
Great ... what do you think about
result2. Now the Indices in the result2 double loop are correct.$endgroup$
– lio
1 hour ago
$begingroup$
Great ... what do you think about
result2. Now the Indices in the result2 double loop are correct.$endgroup$
– lio
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
1 hour ago
add a comment |
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