Substracting previous value in a matrix and assigning it to a new matrix

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Me Me
Me Me 2021-2-4
编辑： Stephen23 2021-2-4
Hi everyone,
I Would like some help doing the following; substracting the previous value in a cell except the first value.
Lets say we have : a=[1 5 3 6 4] , then i would like the output : b= [1 4 2 3 2].
so in the iteration it would be :
  • 1-0 = 1
  • 5-1= 4
  • 3-6 = 3
  • 6-4 = 2
The code works - the only problem is that it does not perform this operation for the first column
a=cell2mat(SortVal)
b=a;
for jj = 2:18
for ii = 2:8
b(jj,ii) = b(jj,ii) - a(jj-1,ii)
end
end
This is a:
a =
1.6766 0.6118 0.3707 0.0771 0.0010 0.2057 0.1219 0.2438
2.2065 0.8679 0.4766 0.0879 0.0017 0.3461 0.1624 0.3007
2.4282 1.0790 0.5340 0.0921 0.0019 0.4416 0.2042 0.3431
2.5295 1.2192 0.5736 0.1017 0.0028 0.4911 0.2265 0.3896
2.6850 1.3342 0.6188 0.1062 0.0032 0.5578 0.2483 0.4327
2.8442 1.3778 0.6719 0.1118 0.0039 0.5993 0.2736 0.4536
3.0451 1.4608 0.7491 0.1133 0.0055 0.6738 0.2959 0.4860
3.1747 1.5049 0.7910 0.1138 0.0062 0.7108 0.3170 0.5029
3.2891 1.5334 0.8172 0.1208 0.0062 0.7247 0.3449 0.5181
3.4065 1.5708 0.8959 0.1238 0.0064 0.7447 0.3641 0.5260
3.4989 1.6455 0.9100 0.1323 0.0064 0.7649 0.3892 0.5480
3.5645 1.6464 0.9129 0.1342 0.0065 0.7828 0.4151 0.5662
3.6503 1.6760 0.9917 0.1364 0.0074 0.7949 0.4171 0.5739
3.7089 1.7567 1.0135 0.1465 0.0077 0.8090 0.4371 0.5834
3.7803 1.7714 1.0432 0.1493 0.0083 0.8243 0.4580 0.5987
3.8278 1.7844 1.0745 0.1540 0.0085 0.8515 0.4608 0.6102
3.8642 1.8361 1.0975 0.1552 0.0114 0.8629 0.4871 0.6260
3.8935 1.8419 1.1220 0.1601 0.0115 0.8665 0.4877 0.6325
and this is my output
b =
1.6766 0.6118 0.3707 0.0771 0.0010 0.2057 0.1219 0.2438
2.2065 0.2560 0.1059 0.0109 0.0007 0.1404 0.0406 0.0569
2.4282 0.2112 0.0574 0.0042 0.0002 0.0955 0.0417 0.0424
2.5295 0.1401 0.0397 0.0096 0.0008 0.0496 0.0224 0.0465
2.6850 0.1150 0.0451 0.0045 0.0004 0.0667 0.0218 0.0431
2.8442 0.0437 0.0532 0.0056 0.0008 0.0415 0.0253 0.0209
3.0451 0.0830 0.0771 0.0014 0.0016 0.0745 0.0222 0.0323
3.1747 0.0441 0.0419 0.0005 0.0007 0.0370 0.0211 0.0170
3.2891 0.0285 0.0262 0.0070 0.0000 0.0139 0.0279 0.0152
3.4065 0.0374 0.0788 0.0030 0.0001 0.0199 0.0192 0.0079
3.4989 0.0747 0.0141 0.0086 0.0000 0.0202 0.0251 0.0219
3.5645 0.0008 0.0029 0.0018 0.0001 0.0179 0.0259 0.0182
3.6503 0.0296 0.0788 0.0022 0.0009 0.0121 0.0020 0.0077
3.7089 0.0808 0.0217 0.0101 0.0003 0.0140 0.0200 0.0095
3.7803 0.0146 0.0297 0.0028 0.0006 0.0153 0.0209 0.0153
3.8278 0.0130 0.0313 0.0048 0.0002 0.0272 0.0028 0.0115
3.8642 0.0517 0.0230 0.0011 0.0029 0.0113 0.0263 0.0158
3.8935 0.0059 0.0244 0.0049 0.0002 0.0036 0.0006 0.0065
The code does it for all the column, but not the first?

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采纳的回答

Stephen23
Stephen23 2021-2-4
Ran in:
a = [
1.6766 0.6118 0.3707 0.0771 0.0010 0.2057 0.1219 0.2438
2.2065 0.8679 0.4766 0.0879 0.0017 0.3461 0.1624 0.3007
2.4282 1.0790 0.5340 0.0921 0.0019 0.4416 0.2042 0.3431
2.5295 1.2192 0.5736 0.1017 0.0028 0.4911 0.2265 0.3896
2.6850 1.3342 0.6188 0.1062 0.0032 0.5578 0.2483 0.4327
2.8442 1.3778 0.6719 0.1118 0.0039 0.5993 0.2736 0.4536
3.0451 1.4608 0.7491 0.1133 0.0055 0.6738 0.2959 0.4860
3.1747 1.5049 0.7910 0.1138 0.0062 0.7108 0.3170 0.5029
3.2891 1.5334 0.8172 0.1208 0.0062 0.7247 0.3449 0.5181
3.4065 1.5708 0.8959 0.1238 0.0064 0.7447 0.3641 0.5260
3.4989 1.6455 0.9100 0.1323 0.0064 0.7649 0.3892 0.5480
3.5645 1.6464 0.9129 0.1342 0.0065 0.7828 0.4151 0.5662
3.6503 1.6760 0.9917 0.1364 0.0074 0.7949 0.4171 0.5739
3.7089 1.7567 1.0135 0.1465 0.0077 0.8090 0.4371 0.5834
3.7803 1.7714 1.0432 0.1493 0.0083 0.8243 0.4580 0.5987
3.8278 1.7844 1.0745 0.1540 0.0085 0.8515 0.4608 0.6102
3.8642 1.8361 1.0975 0.1552 0.0114 0.8629 0.4871 0.6260
3.8935 1.8419 1.1220 0.1601 0.0115 0.8665 0.4877 0.6325];
b = diff(a,1,1)
b = 17×8
0.5299 0.2561 0.1059 0.0108 0.0007 0.1404 0.0405 0.0569 0.2217 0.2111 0.0574 0.0042 0.0002 0.0955 0.0418 0.0424 0.1013 0.1402 0.0396 0.0096 0.0009 0.0495 0.0223 0.0465 0.1555 0.1150 0.0452 0.0045 0.0004 0.0667 0.0218 0.0431 0.1592 0.0436 0.0531 0.0056 0.0007 0.0415 0.0253 0.0209 0.2009 0.0830 0.0772 0.0015 0.0016 0.0745 0.0223 0.0324 0.1296 0.0441 0.0419 0.0005 0.0007 0.0370 0.0211 0.0169 0.1144 0.0285 0.0262 0.0070 0 0.0139 0.0279 0.0152 0.1174 0.0374 0.0787 0.0030 0.0002 0.0200 0.0192 0.0079 0.0924 0.0747 0.0141 0.0085 0 0.0202 0.0251 0.0220
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Me Me
Me Me 2021-2-4
编辑：Stephen23 2021-2-4
Okay, i think i solved it myself now.
I just feel like its not the most effecient way of doing it at all - So if anyone with more programming experience can help - it would be lovely.
Thanks!
a=cell2mat(SortVal)
[r,c]=size(a)
b=a;
for jj = 2:r
b(jj) = b(jj) - a(jj-1)
for ii = 2:c
b(jj,ii) = b(jj,ii) - a(jj-1,ii)
end
end
>> a
a =
1.6766 0.6118 0.3707 0.0771 0.0010 0.2057 0.1219 0.2438
2.2065 0.8679 0.4766 0.0879 0.0017 0.3461 0.1624 0.3007
2.4282 1.0790 0.5340 0.0921 0.0019 0.4416 0.2042 0.3431
2.5295 1.2192 0.5736 0.1017 0.0028 0.4911 0.2265 0.3896
2.6850 1.3342 0.6188 0.1062 0.0032 0.5578 0.2483 0.4327
2.8442 1.3778 0.6719 0.1118 0.0039 0.5993 0.2736 0.4536
3.0451 1.4608 0.7491 0.1133 0.0055 0.6738 0.2959 0.4860
3.1747 1.5049 0.7910 0.1138 0.0062 0.7108 0.3170 0.5029
3.2891 1.5334 0.8172 0.1208 0.0062 0.7247 0.3449 0.5181
3.4065 1.5708 0.8959 0.1238 0.0064 0.7447 0.3641 0.5260
3.4989 1.6455 0.9100 0.1323 0.0064 0.7649 0.3892 0.5480
3.5645 1.6464 0.9129 0.1342 0.0065 0.7828 0.4151 0.5662
3.6503 1.6760 0.9917 0.1364 0.0074 0.7949 0.4171 0.5739
3.7089 1.7567 1.0135 0.1465 0.0077 0.8090 0.4371 0.5834
3.7803 1.7714 1.0432 0.1493 0.0083 0.8243 0.4580 0.5987
3.8278 1.7844 1.0745 0.1540 0.0085 0.8515 0.4608 0.6102
3.8642 1.8361 1.0975 0.1552 0.0114 0.8629 0.4871 0.6260
3.8935 1.8419 1.1220 0.1601 0.0115 0.8665 0.4877 0.6325
and the output:
b =
1.6766 0.6118 0.3707 0.0771 0.0010 0.2057 0.1219 0.2438
0.5299 0.2560 0.1059 0.0109 0.0007 0.1404 0.0406 0.0569
0.2217 0.2112 0.0574 0.0042 0.0002 0.0955 0.0417 0.0424
0.1014 0.1401 0.0397 0.0096 0.0008 0.0496 0.0224 0.0465
0.1555 0.1150 0.0451 0.0045 0.0004 0.0667 0.0218 0.0431
0.1592 0.0437 0.0532 0.0056 0.0008 0.0415 0.0253 0.0209
0.2009 0.0830 0.0771 0.0014 0.0016 0.0745 0.0222 0.0323
0.1296 0.0441 0.0419 0.0005 0.0007 0.0370 0.0211 0.0170
0.1144 0.0285 0.0262 0.0070 0.0000 0.0139 0.0279 0.0152
0.1175 0.0374 0.0788 0.0030 0.0001 0.0199 0.0192 0.0079
0.0923 0.0747 0.0141 0.0086 0.0000 0.0202 0.0251 0.0219
0.0656 0.0008 0.0029 0.0018 0.0001 0.0179 0.0259 0.0182
0.0859 0.0296 0.0788 0.0022 0.0009 0.0121 0.0020 0.0077
0.0585 0.0808 0.0217 0.0101 0.0003 0.0140 0.0200 0.0095
0.0714 0.0146 0.0297 0.0028 0.0006 0.0153 0.0209 0.0153
0.0475 0.0130 0.0313 0.0048 0.0002 0.0272 0.0028 0.0115
0.0363 0.0517 0.0230 0.0011 0.0029 0.0113 0.0263 0.0158
0.0293 0.0059 0.0244 0.0049 0.0002 0.0036 0.0006 0.0065
Stephen23
Stephen23 2021-2-4
编辑：Stephen23 2021-2-4
"I just feel like its not the most effecient way of doing it at all"
Using nested loops is not an efficient approach to this task.
"So if anyone with more programming experience can help - it would be lovely"
As my answer shows, using diff is much simpler and more efficient.
Just concatenate the first line on top and you are done:
b = [a(1,:);diff(a,1,1)]

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