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could anyone tell me how to fix the issue.

1 Ansicht (letzte 30 Tage)
Prabha Kumaresan
Prabha Kumaresan am 26 Jan. 2018
Kommentiert: John D'Errico am 26 Jan. 2018
I am having P_part containing random values of several sizes as seen below
P_part =
Columns 1 through 14
0 0 0.3430 0 0 0 0.2863 0 0.8748 0 0 0 0.1211 0
0 0 0.3127 0 0 0 0.6485 0 0.0673 0 0 0 0.3613 0
0 0 0.5077 0 0 0 0.5520 0 0.8039 0 0 0 0.1442 0
Columns 15 through 18
0 0 0.1138 0
0 0 0.1144 0
0 0 0.6283 0
P_part =
Columns 1 through 14
0.5532 0 0 0 0 0.5641 0 0 0 0 0.3363 0 0 0
0.1395 0 0 0 0 0.5661 0 0 0 0 0.3935 0 0 0
Columns 15 through 18
0 0.0871 0 0
0 0.5110 0 0
P_part =
Columns 1 through 14
0 0 0 0.7541 0 0 0 0 0 0.5487 0 0 0 0.8933
0 0 0 0.3463 0 0 0 0 0 0.9512 0 0 0 0.1732
Columns 15 through 18
0 0 0 0
0 0 0 0
P_part =
Columns 1 through 14
0 0.8925 0 0 0.7796 0 0 0.8006 0 0 0 0.5605 0 0
0 0.3807 0 0 0.2808 0 0 0.2277 0 0 0 0.1999 0 0
0 0.9237 0 0 0.2959 0 0 0.1927 0 0 0 0.6744 0 0
Columns 15 through 18
0.4589 0 0 0.7834
0.2623 0 0 0.2107
0.2719 0 0 0.7777
P_part =
Columns 1 through 14
0.4163 0 0.5580 0 0 0 0 0 0 0 0.6975 0 0.1935 0
0.3575 0 0.5808 0 0 0 0 0 0 0 0.4303 0 0.6271 0
Columns 15 through 20
0 0 0 0 0 0
0 0 0 0 0 0
P_part =
Columns 1 through 14
0 0.0274 0 0 0 0 0.0455 0 0 0.7313 0 0.3900 0 0
0 0.7636 0 0 0 0 0.0393 0 0 0.9110 0 0.7803 0 0
0 0.9306 0 0 0 0 0.9057 0 0 0.2264 0 0.0977 0 0
Columns 15 through 20
0 0 0.6932 0 0 0.3759
0 0 0.3161 0 0 0.7273
0 0 0.8646 0 0 0.0428
P_part =
Columns 1 through 14
0 0 0 0 0.8811 0.8666 0 0.6317 0 0 0 0 0 0
0 0 0 0 0.5550 0.6808 0 0.0659 0 0 0 0 0 0
0 0 0 0 0.1760 0.8051 0 0.5753 0 0 0 0 0 0
Columns 15 through 20
0.7551 0.4270 0 0.9280 0 0
0.6956 0.5850 0 0.4978 0 0
0.2394 0.4494 0 0.6998 0 0
P_part =
Columns 1 through 14
0 0 0 0.2330 0 0 0 0 0.7534 0 0 0 0 0.5109
0 0 0 0.3144 0 0 0 0 0.6663 0 0 0 0 0.7208
Columns 15 through 20
0 0 0 0 0.0436 0
0 0 0 0 0.1191 0
could anyone tell me how to have the random values within the interval [0.01,0.09] such that the sum of the values of each P_part should be same. could anyone help me how to obtain it.
  1 Kommentar
John D'Errico
John D'Errico am 26 Jan. 2018
This is now what, the 5th time you have asked this same question? STOP THE REPEAT POSTINGS.

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