Data coverage per cell of 3D array

2022 4 8
1 个回答
回答已采纳
更新时间：2022 4 9
5 次查看（30 天）

0 个投票

I have a 3D data array of size 150 x 75 x 730 (lat x lon x time (days). Data coverage for each day (3rd dimension) varies with NAN values representing no data.
How do I calculate the data coverage in each cell for the total time duration? I want to use this to select cells with suffiicient data coverage.

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Lisa Hancke
Lisa Hancke 2022-4-9
编辑：Lisa Hancke 2022-4-9
Thanks. I used
isfinite
and then
sum
along the 3rd dimension to get the same result. Easy :)

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Voss
Voss 2022-4-8
Ran in:

0 个投票

Ran in:
% create a 3D matrix data with 15000 elements
data = rand(15,10,100);
% put NaNs at 7500 random indices (which may have repeats)
data(randi(numel(data),7500,1)) = NaN;
% this many elements of data are NaN:
nnz(isnan(data))
ans = 5918
% data coverage is the total number of non-NaN elements
% at each (x,y) point, where x and y correspond to the
% first 2 dimensions of data.
% sum ~isnan(data) over the third dimension to find
% the data coverage:
data_coverage = sum(~isnan(data),3);
disp(data_coverage);
61 61 63 61 67 61 56 57 61 69 67 62 59 67 62 59 60 56 62 56 57 69 61 60 60 65 61 66 65 58 60 68 58 62 64 52 57 61 62 55 65 55 58 60 69 61 65 57 55 60 59 54 68 52 60 64 60 55 63 58 64 54 65 63 59 56 62 61 46 58 68 55 60 64 63 58 63 63 55 58 54 55 61 59 69 62 60 49 61 60 59 61 63 64 57 63 63 58 73 55 64 60 61 62 57 60 65 60 56 67 52 65 50 53 67 65 62 54 55 64 61 68 61 58 68 61 69 56 62 62 52 65 60 63 56 67 57 64 62 67 62 56 68 58 60 69 57 58 51 58
% say sufficient data coverage is >= 60 non-NaNs
has_sufficient_data = data_coverage >= 60 % using the number of non-NaNs
has_sufficient_data = 15×10 logical array
1 1 1 1 1 1 0 0 1 1 1 1 0 1 1 0 1 0 1 0 0 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 0 1 1 0 0 0 0
% which is the same as <= 40 NaNs:
has_sufficient_data = sum(isnan(data),3) <= 40 % using the number of NaNs
has_sufficient_data = 15×10 logical array
1 1 1 1 1 1 0 0 1 1 1 1 0 1 1 0 1 0 1 0 0 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 0 1 1 0 0 0 0

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 NaNs 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by