Counting certain elements in lists
5
$begingroup$
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:
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 yesterday
liolio
1,147217
$endgroup$
add a comment |
5
$begingroup$
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:
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 yesterday
liolio
1,147217
$endgroup$
add a comment |
5
5
5
$begingroup$
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:
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 yesterday
liolio
1,147217
$endgroup$
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:
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 yesterday
liolio
1,147217
asked yesterday
liolio
1,147217
edited 21 hours ago
lio
asked yesterday
liolio
1,147217
asked yesterday
liolio
1,147217
asked yesterday
liolio
1,147217
1,147217
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
4
$begingroup$
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}
Timings: Timing comparison with OP's Do loop and MarcoB's Map[Count,...] for data2:
n1 = 1000;
n3 = 50;
data2 = Array[0 &, n3];
SeedRandom[1]
Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];
result2a = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}]; //
RepeatedTiming // First
0.037
result2b = Map[Count[1], data2[[All, All, 2, 1]], {2}]; // RepeatedTiming // First
0.12
(result2 = Array[0 &, n3];
Do[result2[[j]] = Table[Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
{i, 1, n1}];, {j, 1, n3}];) // RepeatedTiming // First
7.8
result2 == result2a == result2b
True
answered yesterday
kglrkglr
189k10205422
$endgroup$
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
Since especiallyn3in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation ofdata2also be expressed instead of using the do loopDo[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?
$endgroup$
– lio
17 hours ago
1
$begingroup$
@li, thank you for the accept. You can dodata2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]
$endgroup$
– kglr
17 hours ago
add a comment |
5
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], {2}];
answered yesterday
MarcoBMarcoB
37.5k556113
$endgroup$
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
yesterday
1
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
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}
Timings: Timing comparison with OP's Do loop and MarcoB's Map[Count,...] for data2:
n1 = 1000;
n3 = 50;
data2 = Array[0 &, n3];
SeedRandom[1]
Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];
result2a = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}]; //
RepeatedTiming // First
0.037
result2b = Map[Count[1], data2[[All, All, 2, 1]], {2}]; // RepeatedTiming // First
0.12
(result2 = Array[0 &, n3];
Do[result2[[j]] = Table[Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
{i, 1, n1}];, {j, 1, n3}];) // RepeatedTiming // First
7.8
result2 == result2a == result2b
True
answered yesterday
kglrkglr
189k10205422
$endgroup$
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
Since especiallyn3in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation ofdata2also be expressed instead of using the do loopDo[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?
$endgroup$
– lio
17 hours ago
1
$begingroup$
@li, thank you for the accept. You can dodata2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]
$endgroup$
– kglr
17 hours ago
add a comment |
4
$begingroup$
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}
Timings: Timing comparison with OP's Do loop and MarcoB's Map[Count,...] for data2:
n1 = 1000;
n3 = 50;
data2 = Array[0 &, n3];
SeedRandom[1]
Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];
result2a = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}]; //
RepeatedTiming // First
0.037
result2b = Map[Count[1], data2[[All, All, 2, 1]], {2}]; // RepeatedTiming // First
0.12
(result2 = Array[0 &, n3];
Do[result2[[j]] = Table[Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
{i, 1, n1}];, {j, 1, n3}];) // RepeatedTiming // First
7.8
result2 == result2a == result2b
True
answered yesterday
kglrkglr
189k10205422
$endgroup$
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
Since especiallyn3in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation ofdata2also be expressed instead of using the do loopDo[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?
$endgroup$
– lio
17 hours ago
1
$begingroup$
@li, thank you for the accept. You can dodata2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]
$endgroup$
– kglr
17 hours ago
add a comment |
4
4
4
$begingroup$
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}
Timings: Timing comparison with OP's Do loop and MarcoB's Map[Count,...] for data2:
n1 = 1000;
n3 = 50;
data2 = Array[0 &, n3];
SeedRandom[1]
Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];
result2a = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}]; //
RepeatedTiming // First
0.037
result2b = Map[Count[1], data2[[All, All, 2, 1]], {2}]; // RepeatedTiming // First
0.12
(result2 = Array[0 &, n3];
Do[result2[[j]] = Table[Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
{i, 1, n1}];, {j, 1, n3}];) // RepeatedTiming // First
7.8
result2 == result2a == result2b
True
answered yesterday
kglrkglr
189k10205422
$endgroup$
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}
Timings: Timing comparison with OP's Do loop and MarcoB's Map[Count,...] for data2:
n1 = 1000;
n3 = 50;
data2 = Array[0 &, n3];
SeedRandom[1]
Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];
result2a = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], {4}]; //
RepeatedTiming // First
0.037
result2b = Map[Count[1], data2[[All, All, 2, 1]], {2}]; // RepeatedTiming // First
0.12
(result2 = Array[0 &, n3];
Do[result2[[j]] = Table[Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
{i, 1, n1}];, {j, 1, n3}];) // RepeatedTiming // First
7.8
result2 == result2a == result2b
True
answered yesterday
kglrkglr
189k10205422
edited 21 hours ago
answered yesterday
kglrkglr
189k10205422
answered yesterday
kglrkglr
189k10205422
answered yesterday
kglrkglr
189k10205422
189k10205422
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
Since especiallyn3in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation ofdata2also be expressed instead of using the do loopDo[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?
$endgroup$
– lio
17 hours ago
1
$begingroup$
@li, thank you for the accept. You can dodata2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]
$endgroup$
– kglr
17 hours ago
add a comment |
$begingroup$
Great ... what do you think aboutresult2. Now the Indices in theresult2double loop are correct.
$endgroup$
– lio
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
Since especiallyn3in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation ofdata2also be expressed instead of using the do loopDo[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?
$endgroup$
– lio
17 hours ago
1
$begingroup$
@li, thank you for the accept. You can dodata2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]
$endgroup$
– kglr
17 hours ago
$begingroup$
Great ... what do you think about
result2. Now the Indices in the result2 double loop are correct.$endgroup$
– lio
yesterday
$begingroup$
Great ... what do you think about
result2. Now the Indices in the result2 double loop are correct.$endgroup$
– lio
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
@lio, please see the update.
$endgroup$
– kglr
yesterday
$begingroup$
Since especially
n3 in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation of data2 also be expressed instead of using the do loop Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?$endgroup$
– lio
17 hours ago
$begingroup$
Since especially
n3 in my real problems is much larger than 50 I vote for your fast solution. Thank you for the timings. May I ask you something: how can the creation of data2 also be expressed instead of using the do loop Do[data2[[i]] = {#, {RandomInteger[3, 20]}} & /@ Range[n1];, {i, 1, n3}];?$endgroup$
– lio
17 hours ago
1
1
$begingroup$
@li, thank you for the accept. You can do
data2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]$endgroup$
– kglr
17 hours ago
$begingroup$
@li, thank you for the accept. You can do
data2b = Table[ Transpose[{Range[n1], RandomInteger[3, {n1, 20}]}], {i, 1, n3}]$endgroup$
– kglr
17 hours ago
add a comment |
5
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], {2}];
answered yesterday
MarcoBMarcoB
37.5k556113
$endgroup$
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
yesterday
1
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
yesterday
add a comment |
5
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], {2}];
answered yesterday
MarcoBMarcoB
37.5k556113
$endgroup$
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
yesterday
1
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
yesterday
add a comment |
5
5
5
$begingroup$
For your result1:
Count[1] /@ data1[[All, 2, 1]]
For your result2:
Map[Count[1], data2[[All, All, 2, 1]], {2}];
answered yesterday
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 yesterday
MarcoBMarcoB
37.5k556113
edited yesterday
answered yesterday
MarcoBMarcoB
37.5k556113
answered yesterday
MarcoBMarcoB
37.5k556113
answered yesterday
MarcoBMarcoB
37.5k556113
37.5k556113
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
yesterday
1
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
yesterday
add a comment |
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea forresult2improvement?
$endgroup$
– lio
yesterday
1
$begingroup$
@lio Yes, I've added something for result2 as well. You can check that they give the same output as yourresult1andresult2respectively.
$endgroup$
– MarcoB
yesterday
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
This is great ...
$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea for
result2 improvement?$endgroup$
– lio
yesterday
$begingroup$
Do you have an idea for
result2 improvement?$endgroup$
– lio
yesterday
1
1
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
add a comment |
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