Averaging over columns while ignoring zero entries
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:
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];
result // N
{2., 3., 3., 4., 4., 3., 9.}
How can I replace the Do loop?
list-manipulation
edited 12 hours ago
m_goldberg
87.4k872198
asked 15 hours ago
liolio
1,108217
$endgroup$
add a 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:
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];
result // N
{2., 3., 3., 4., 4., 3., 9.}
How can I replace the Do loop?
list-manipulation
edited 12 hours ago
m_goldberg
87.4k872198
asked 15 hours ago
liolio
1,108217
$endgroup$
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
15 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
15 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
15 hours ago
4
$begingroup$
I'm not at a computer right now, so please tryTotal[list]/Total[Unitize[list]].
$endgroup$
– J. M. is computer-less♦
14 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
12 hours ago
add a 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:
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];
result // N
{2., 3., 3., 4., 4., 3., 9.}
How can I replace the Do loop?
list-manipulation
edited 12 hours ago
m_goldberg
87.4k872198
asked 15 hours ago
liolio
1,108217
$endgroup$
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:
Do[
array = list[[All, i]];
total = Total[array];
cnt = Count[array, 0];
result[[i]] = total/(Length@array - cnt);,
{i, 1, Length@result}
];
result // N
{2., 3., 3., 4., 4., 3., 9.}
How can I replace the Do loop?
list-manipulation
list-manipulation
edited 12 hours ago
m_goldberg
87.4k872198
asked 15 hours ago
liolio
1,108217
edited 12 hours ago
m_goldberg
87.4k872198
asked 15 hours ago
liolio
1,108217
edited 12 hours ago
m_goldberg
87.4k872198
edited 12 hours ago
m_goldberg
87.4k872198
edited 12 hours ago
m_goldberg
87.4k872198
87.4k872198
asked 15 hours ago
liolio
1,108217
asked 15 hours ago
liolio
1,108217
asked 15 hours ago
liolio
1,108217
1,108217
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
15 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
15 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
15 hours ago
4
$begingroup$
I'm not at a computer right now, so please tryTotal[list]/Total[Unitize[list]].
$endgroup$
– J. M. is computer-less♦
14 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
12 hours ago
add a comment |
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
15 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
15 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
15 hours ago
4
$begingroup$
I'm not at a computer right now, so please tryTotal[list]/Total[Unitize[list]].
$endgroup$
– J. M. is computer-less♦
14 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
12 hours ago
1
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]$endgroup$
– MarcoB
15 hours ago
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]$endgroup$
– MarcoB
15 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
15 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
15 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
15 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
15 hours ago
4
4
$begingroup$
I'm not at a computer right now, so please try
Total[list]/Total[Unitize[list]].$endgroup$
– J. M. is computer-less♦
14 hours ago
$begingroup$
I'm not at a computer right now, so please try
Total[list]/Total[Unitize[list]].$endgroup$
– J. M. is computer-less♦
14 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
12 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
12 hours ago
add a comment |
2 Answers
2
active
oldest
votes
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
{2,3,3,4,4,3,9}
answered 15 hours ago
MarcoBMarcoB
37.2k556113
$endgroup$
add a comment |
7
$begingroup$
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}
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
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votes
active
oldest
votes
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
{2,3,3,4,4,3,9}
answered 15 hours ago
MarcoBMarcoB
37.2k556113
$endgroup$
add a comment |
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
{2,3,3,4,4,3,9}
answered 15 hours ago
MarcoBMarcoB
37.2k556113
$endgroup$
add a comment |
8
8
8
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
{2,3,3,4,4,3,9}
answered 15 hours ago
MarcoBMarcoB
37.2k556113
$endgroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
{2,3,3,4,4,3,9}
answered 15 hours ago
MarcoBMarcoB
37.2k556113
answered 15 hours ago
MarcoBMarcoB
37.2k556113
answered 15 hours ago
MarcoBMarcoB
37.2k556113
answered 15 hours ago
MarcoBMarcoB
37.2k556113
37.2k556113
add a comment |
add a comment |
7
$begingroup$
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}
$endgroup$
add a comment |
7
$begingroup$
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}
$endgroup$
add a comment |
7
7
7
$begingroup$
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}
$endgroup$
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}
answered 12 hours ago
community wiki
m_goldberg
add a comment |
add a comment |
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1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]$endgroup$
– MarcoB
15 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
15 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
15 hours ago
4
$begingroup$
I'm not at a computer right now, so please try
Total[list]/Total[Unitize[list]].$endgroup$
– J. M. is computer-less♦
14 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
12 hours ago