Averaging over columns while ignoring zero entries
Clash Royale CLAN TAG#URR8PPP
5
$begingroup$
I have:
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0,
1, 0, 2, 2, 2, 4, 0
I want to calculate the average but without considering the 0s.
The result should be:
result = Array[0 &, Last@Dimensions@list];
n = Length@result;
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
If[Length@array - cnt > 0,
result[[i]] = total/(Length@array - cnt),
result[[i]] = 0
];,
i, 1, n
];
result // N
2., 3., 3., 4., 4., 3., 9.
How can I replace the Do loop?
list-manipulation
asked Mar 8 at 13:50
liolio
1,102218
$endgroup$
|
show 1 more comment
5
$begingroup$
I have:
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0,
1, 0, 2, 2, 2, 4, 0
I want to calculate the average but without considering the 0s.
The result should be:
result = Array[0 &, Last@Dimensions@list];
n = Length@result;
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
If[Length@array - cnt > 0,
result[[i]] = total/(Length@array - cnt),
result[[i]] = 0
];,
i, 1, n
];
result // N
2., 3., 3., 4., 4., 3., 9.
How can I replace the Do loop?
list-manipulation
asked Mar 8 at 13:50
liolio
1,102218
$endgroup$
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
Mar 8 at 14:10
$begingroup$
Thank you for the solution.
$endgroup$
– lio
Mar 8 at 14:12
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
Mar 8 at 14:13
5
$begingroup$
I'm not at a computer right now, so please tryTotal[list]/Total[Unitize[list]].
$endgroup$
– J. M. is away♦
Mar 8 at 14:21
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
Mar 8 at 16:50
|
show 1 more comment
5
5
5
$begingroup$
I have:
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0,
1, 0, 2, 2, 2, 4, 0
I want to calculate the average but without considering the 0s.
The result should be:
result = Array[0 &, Last@Dimensions@list];
n = Length@result;
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
If[Length@array - cnt > 0,
result[[i]] = total/(Length@array - cnt),
result[[i]] = 0
];,
i, 1, n
];
result // N
2., 3., 3., 4., 4., 3., 9.
How can I replace the Do loop?
list-manipulation
asked Mar 8 at 13:50
liolio
1,102218
$endgroup$
I have:
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0,
1, 0, 2, 2, 2, 4, 0
I want to calculate the average but without considering the 0s.
The result should be:
result = Array[0 &, Last@Dimensions@list];
n = Length@result;
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
If[Length@array - cnt > 0,
result[[i]] = total/(Length@array - cnt),
result[[i]] = 0
];,
i, 1, n
];
result // N
2., 3., 3., 4., 4., 3., 9.
How can I replace the Do loop?
list-manipulation
list-manipulation
asked Mar 8 at 13:50
liolio
1,102218
asked Mar 8 at 13:50
liolio
1,102218
edited Mar 11 at 15:06
lio
asked Mar 8 at 13:50
liolio
1,102218
asked Mar 8 at 13:50
liolio
1,102218
asked Mar 8 at 13:50
liolio
1,102218
1,102218
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
Mar 8 at 14:10
$begingroup$
Thank you for the solution.
$endgroup$
– lio
Mar 8 at 14:12
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
Mar 8 at 14:13
5
$begingroup$
I'm not at a computer right now, so please tryTotal[list]/Total[Unitize[list]].
$endgroup$
– J. M. is away♦
Mar 8 at 14:21
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
Mar 8 at 16:50
|
show 1 more comment
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
Mar 8 at 14:10
$begingroup$
Thank you for the solution.
$endgroup$
– lio
Mar 8 at 14:12
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
Mar 8 at 14:13
5
$begingroup$
I'm not at a computer right now, so please tryTotal[list]/Total[Unitize[list]].
$endgroup$
– J. M. is away♦
Mar 8 at 14:21
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
Mar 8 at 16:50
1
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]$endgroup$
– MarcoB
Mar 8 at 14:10
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]$endgroup$
– MarcoB
Mar 8 at 14:10
$begingroup$
Thank you for the solution.
$endgroup$
– lio
Mar 8 at 14:12
$begingroup$
Thank you for the solution.
$endgroup$
– lio
Mar 8 at 14:12
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
Mar 8 at 14:13
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
Mar 8 at 14:13
5
5
$begingroup$
I'm not at a computer right now, so please try
Total[list]/Total[Unitize[list]].$endgroup$
– J. M. is away♦
Mar 8 at 14:21
$begingroup$
I'm not at a computer right now, so please try
Total[list]/Total[Unitize[list]].$endgroup$
– J. M. is away♦
Mar 8 at 14:21
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
Mar 8 at 16:50
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
Mar 8 at 16:50
|
show 1 more comment
3 Answers
3
active
oldest
votes
8
$begingroup$
I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/Total[Unitize[list]]
2, 3, 3, 4, 4, 3, 9
Update
As J.M. points out in a comment below, there is more robust formulation that handles the cases where one or mort columns contain all zeros.
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/(Total[Unitize[list]] /. 0 -> 1)
2, 3, 3, 4, 4, 3, 0
$endgroup$
1
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
3
$begingroup$
@lio, change the denominator to(Total[Unitize[list]] /. 0 -> 1).
$endgroup$
– J. M. is away♦
Mar 11 at 15:02
add a comment |
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
$endgroup$
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
@lio, It does not produce an error. It just returnsMean[]in the case that all entries are zero. What result would you expect in that case?
$endgroup$
– MarcoB
Mar 11 at 16:09
1
$begingroup$
I think the OP expects2, 3, 3, 4, 4, 3, 0in this case.
$endgroup$
– m_goldberg
Mar 11 at 18:20
add a comment |
1
$begingroup$
ClearAll[means]
means = Mean[#["NonzeroValues"] /. -> 0] & /@ SparseArray[Transpose[#]] &;
Examples:
list1 = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list1
2, 3, 3, 4, 4, 3, 9
list2 = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list2
2, 3, 3, 4, 4, 3, 0
answered Mar 11 at 20:54
kglrkglr
190k10206425
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
$begingroup$
I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/Total[Unitize[list]]
2, 3, 3, 4, 4, 3, 9
Update
As J.M. points out in a comment below, there is more robust formulation that handles the cases where one or mort columns contain all zeros.
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/(Total[Unitize[list]] /. 0 -> 1)
2, 3, 3, 4, 4, 3, 0
$endgroup$
1
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
3
$begingroup$
@lio, change the denominator to(Total[Unitize[list]] /. 0 -> 1).
$endgroup$
– J. M. is away♦
Mar 11 at 15:02
add a comment |
8
$begingroup$
I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/Total[Unitize[list]]
2, 3, 3, 4, 4, 3, 9
Update
As J.M. points out in a comment below, there is more robust formulation that handles the cases where one or mort columns contain all zeros.
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/(Total[Unitize[list]] /. 0 -> 1)
2, 3, 3, 4, 4, 3, 0
$endgroup$
1
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
3
$begingroup$
@lio, change the denominator to(Total[Unitize[list]] /. 0 -> 1).
$endgroup$
– J. M. is away♦
Mar 11 at 15:02
add a comment |
8
8
8
$begingroup$
I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/Total[Unitize[list]]
2, 3, 3, 4, 4, 3, 9
Update
As J.M. points out in a comment below, there is more robust formulation that handles the cases where one or mort columns contain all zeros.
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/(Total[Unitize[list]] /. 0 -> 1)
2, 3, 3, 4, 4, 3, 0
$endgroup$
I'm posting this as a CW answer, so J.M.'s very nice answer, made in a comment above, gets recorded as a real answer.
list = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/Total[Unitize[list]]
2, 3, 3, 4, 4, 3, 9
Update
As J.M. points out in a comment below, there is more robust formulation that handles the cases where one or mort columns contain all zeros.
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
Total[list]/(Total[Unitize[list]] /. 0 -> 1)
2, 3, 3, 4, 4, 3, 0
edited Mar 11 at 18:24
community wiki
2 revs
m_goldberg
1
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
3
$begingroup$
@lio, change the denominator to(Total[Unitize[list]] /. 0 -> 1).
$endgroup$
– J. M. is away♦
Mar 11 at 15:02
add a comment |
1
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
3
$begingroup$
@lio, change the denominator to(Total[Unitize[list]] /. 0 -> 1).
$endgroup$
– J. M. is away♦
Mar 11 at 15:02
1
1
$begingroup$
This produces an error for:
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
This produces an error for:
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0$endgroup$
– lio
Mar 11 at 14:58
3
3
$begingroup$
@lio, change the denominator to
(Total[Unitize[list]] /. 0 -> 1).$endgroup$
– J. M. is away♦
Mar 11 at 15:02
$begingroup$
@lio, change the denominator to
(Total[Unitize[list]] /. 0 -> 1).$endgroup$
– J. M. is away♦
Mar 11 at 15:02
add a comment |
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
$endgroup$
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
@lio, It does not produce an error. It just returnsMean[]in the case that all entries are zero. What result would you expect in that case?
$endgroup$
– MarcoB
Mar 11 at 16:09
1
$begingroup$
I think the OP expects2, 3, 3, 4, 4, 3, 0in this case.
$endgroup$
– m_goldberg
Mar 11 at 18:20
add a comment |
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
$endgroup$
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
@lio, It does not produce an error. It just returnsMean[]in the case that all entries are zero. What result would you expect in that case?
$endgroup$
– MarcoB
Mar 11 at 16:09
1
$begingroup$
I think the OP expects2, 3, 3, 4, 4, 3, 0in this case.
$endgroup$
– m_goldberg
Mar 11 at 18:20
add a comment |
8
8
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
$endgroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
answered Mar 8 at 14:12
MarcoBMarcoB
38.5k557115
38.5k557115
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
@lio, It does not produce an error. It just returnsMean[]in the case that all entries are zero. What result would you expect in that case?
$endgroup$
– MarcoB
Mar 11 at 16:09
1
$begingroup$
I think the OP expects2, 3, 3, 4, 4, 3, 0in this case.
$endgroup$
– m_goldberg
Mar 11 at 18:20
add a comment |
$begingroup$
This produces an error for:list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0
$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
@lio, It does not produce an error. It just returnsMean[]in the case that all entries are zero. What result would you expect in that case?
$endgroup$
– MarcoB
Mar 11 at 16:09
1
$begingroup$
I think the OP expects2, 3, 3, 4, 4, 3, 0in this case.
$endgroup$
– m_goldberg
Mar 11 at 18:20
$begingroup$
This produces an error for:
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
This produces an error for:
list = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0$endgroup$
– lio
Mar 11 at 14:58
$begingroup$
@lio, It does not produce an error. It just returns
Mean[] in the case that all entries are zero. What result would you expect in that case?$endgroup$
– MarcoB
Mar 11 at 16:09
$begingroup$
@lio, It does not produce an error. It just returns
Mean[] in the case that all entries are zero. What result would you expect in that case?$endgroup$
– MarcoB
Mar 11 at 16:09
1
1
$begingroup$
I think the OP expects
2, 3, 3, 4, 4, 3, 0 in this case.$endgroup$
– m_goldberg
Mar 11 at 18:20
$begingroup$
I think the OP expects
2, 3, 3, 4, 4, 3, 0 in this case.$endgroup$
– m_goldberg
Mar 11 at 18:20
add a comment |
1
$begingroup$
ClearAll[means]
means = Mean[#["NonzeroValues"] /. -> 0] & /@ SparseArray[Transpose[#]] &;
Examples:
list1 = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list1
2, 3, 3, 4, 4, 3, 9
list2 = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list2
2, 3, 3, 4, 4, 3, 0
answered Mar 11 at 20:54
kglrkglr
190k10206425
$endgroup$
add a comment |
1
$begingroup$
ClearAll[means]
means = Mean[#["NonzeroValues"] /. -> 0] & /@ SparseArray[Transpose[#]] &;
Examples:
list1 = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list1
2, 3, 3, 4, 4, 3, 9
list2 = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list2
2, 3, 3, 4, 4, 3, 0
answered Mar 11 at 20:54
kglrkglr
190k10206425
$endgroup$
add a comment |
1
1
1
$begingroup$
ClearAll[means]
means = Mean[#["NonzeroValues"] /. -> 0] & /@ SparseArray[Transpose[#]] &;
Examples:
list1 = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list1
2, 3, 3, 4, 4, 3, 9
list2 = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list2
2, 3, 3, 4, 4, 3, 0
answered Mar 11 at 20:54
kglrkglr
190k10206425
$endgroup$
ClearAll[means]
means = Mean[#["NonzeroValues"] /. -> 0] & /@ SparseArray[Transpose[#]] &;
Examples:
list1 = 1, 3, 4, 5, 6, 0, 9, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list1
2, 3, 3, 4, 4, 3, 9
list2 = 1, 3, 4, 5, 6, 0, 0, 4, 0, 3, 5, 0, 2, 0, 1, 0, 2, 2, 2, 4, 0;
means @ list2
2, 3, 3, 4, 4, 3, 0
answered Mar 11 at 20:54
kglrkglr
190k10206425
answered Mar 11 at 20:54
kglrkglr
190k10206425
answered Mar 11 at 20:54
kglrkglr
190k10206425
answered Mar 11 at 20:54
kglrkglr
190k10206425
190k10206425
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1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]$endgroup$
– MarcoB
Mar 8 at 14:10
$begingroup$
Thank you for the solution.
$endgroup$
– lio
Mar 8 at 14:12
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
Mar 8 at 14:13
5
$begingroup$
I'm not at a computer right now, so please try
Total[list]/Total[Unitize[list]].$endgroup$
– J. M. is away♦
Mar 8 at 14:21
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
Mar 8 at 16:50