Average every 3 rows of 1 column in a 12 x 8 array

I have a 12 x 8 array of data and want to average every 3 rows of the 4th column in the array, resulting in a 4 x 1 array. To be extremely clear, I want rows 1 to 3 averaged, then 4 to 6 averaged, then 7 to 9 averaged then 10 to 12 averaged.

I need to also apply this to other arrays that have different numbers of rows but still in multiples of 3 (e.g., could be 15 x 8 or 18 x 8 swapped for the original 12 x 8).

%Here is the example data in the 12 x 8 array

34	6	4	-6.60874766440390	-40.7725049965035	16217	0.289000000000000	1.02200000000000
35	6	5	-6.54326464638770	-40.5611336664881	16525	0.286000000000000	1.23700000000000
36	6	6	-6.54628693952691	-40.6729573224369	16386	0.202000000000000	1.13000000000000
172	6	4	-6.61176995754311	-39.6642486096006	16209	0.232000000000000	1.03300000000000
173	6	5	-6.62184426800714	-40.0039787602111	16212	0.236000000000000	1.02700000000000
174	6	6	-6.66113407881686	-39.9370137005201	16069	0.271000000000000	1.16200000000000
310	6	4	-6.64803747521362	-40.3153910672626	16143	0.239000000000000	1.13100000000000
311	6	5	-6.61479225068232	-40.5411679705423	16113	0.285000000000000	1.06400000000000
312	6	6	-6.63695573370318	-40.0871548121586	16266	0.249000000000000	1.02200000000000
448	6	4	-6.76792176973558	-40.3663142652769	16176	0.297000000000000	1.01800000000000
449	6	5	-6.68631985497693	-40.6696972594111	16075	0.277000000000000	1.19400000000000
450	6	6	-6.68631985497693	-40.8846682513060	16046	0.238000000000000	1.17900000000000

[VBBV]

Data = [
34	6	4	-6.60874766440390	-40.7725049965035	16217	0.289000000000000	1.02200000000000
35	6	5	-6.54326464638770	-40.5611336664881	16525	0.286000000000000	1.23700000000000
36	6	6	-6.54628693952691	-40.6729573224369	16386	0.202000000000000	1.13000000000000
172	6	4	-6.61176995754311	-39.6642486096006	16209	0.232000000000000	1.03300000000000
173	6	5	-6.62184426800714	-40.0039787602111	16212	0.236000000000000	1.02700000000000
174	6	6	-6.66113407881686	-39.9370137005201	16069	0.271000000000000	1.16200000000000
310	6	4	-6.64803747521362	-40.3153910672626	16143	0.239000000000000	1.13100000000000
311	6	5	-6.61479225068232	-40.5411679705423	16113	0.285000000000000	1.06400000000000
312	6	6	-6.63695573370318	-40.0871548121586	16266	0.249000000000000	1.02200000000000
448	6	4	-6.76792176973558	-40.3663142652769	16176	0.297000000000000	1.01800000000000
449	6	5	-6.68631985497693	-40.6696972594111	16075	0.277000000000000	1.19400000000000
450	6	6	-6.68631985497693	-40.8846682513060	16046	0.238000000000000	1.17900000000000];
K = 1;
for k = 1:4
   Mean(k) = mean(Data(K:K+2,4));
   K = K + 3;
end
Mean.'

ans = 4×1

-6.5661 -6.6316 -6.6333 -6.7135

Answer 1

[James Tursa]

E.g.,

Data = [
    34	6	4	-6.60874766440390	-40.7725049965035	16217	0.289000000000000	1.02200000000000
    35	6	5	-6.54326464638770	-40.5611336664881	16525	0.286000000000000	1.23700000000000
    36	6	6	-6.54628693952691	-40.6729573224369	16386	0.202000000000000	1.13000000000000
    172	6	4	-6.61176995754311	-39.6642486096006	16209	0.232000000000000	1.03300000000000
    173	6	5	-6.62184426800714	-40.0039787602111	16212	0.236000000000000	1.02700000000000
    174	6	6	-6.66113407881686	-39.9370137005201	16069	0.271000000000000	1.16200000000000
    310	6	4	-6.64803747521362	-40.3153910672626	16143	0.239000000000000	1.13100000000000
    311	6	5	-6.61479225068232	-40.5411679705423	16113	0.285000000000000	1.06400000000000
    312	6	6	-6.63695573370318	-40.0871548121586	16266	0.249000000000000	1.02200000000000
    448	6	4	-6.76792176973558	-40.3663142652769	16176	0.297000000000000	1.01800000000000
    449	6	5	-6.68631985497693	-40.6696972594111	16075	0.277000000000000	1.19400000000000
    450	6	6	-6.68631985497693	-40.8846682513060	16046	0.238000000000000	1.17900000000000];

mean(reshape(Data(:,4),3,[])).'

ans = 4×1

-6.5661 -6.6316 -6.6333 -6.7135

The reshape( ) puts the data you want averaged into individual columns of three elements each, and the mean( ) automatically operates on columns giving the result you want. This naturally results in a row vector, so the final transpose reshapes it into a column vector. The method shown works for any sized matrix where the number of rows is a multiple of 3.

[Emily]

thank you!! very clear explanation, much appreciated