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Averaging over columns while ignoring zero entries
Operations on a List of Vectors (Norm)Compute percentages in a listElegant way to map a function over the columns of an arrayHow to count number of observations/rows within each groupHow can Tally group elements that differ only by a sign together?Creating Average Grid data from Co-ordinate and value dataList manipulation, averaging over lists and fidelityHow to select elements from a list of pairs based on 2nd element of the pairSubtract value from listHow to list ArrayFilter?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 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♦
13 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
11 hours ago
add a comment |
$begingroup$
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
$endgroup$
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 11 hours ago
m_goldberg
87.4k872198
87.4k872198
asked 14 hours ago
liolio
1,108217
1,108217
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 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♦
13 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
11 hours ago
add a comment |
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 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♦
13 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
11 hours ago
1
1
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
$endgroup$
– MarcoB
13 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 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♦
13 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♦
13 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
11 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
$endgroup$
add a comment |
$begingroup$
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 |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
$endgroup$
add a comment |
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
$endgroup$
add a comment |
$begingroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
$endgroup$
Mean /@ DeleteCases[Transpose@list, 0, All]
2,3,3,4,4,3,9
answered 13 hours ago
MarcoBMarcoB
37.2k556113
37.2k556113
add a comment |
add a comment |
$begingroup$
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 |
$begingroup$
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 |
$begingroup$
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 11 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
13 hours ago
$begingroup$
Thank you for the solution.
$endgroup$
– lio
13 hours ago
$begingroup$
You’re welcome. I’ve added an answer since this solution works for you
$endgroup$
– MarcoB
13 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♦
13 hours ago
$begingroup$
@J.M. Well, that is clever! Cool trick :-)
$endgroup$
– MarcoB
11 hours ago