Cody

# Problem 71. Read a column of numbers and interpolate missing data

Solution 1799574

Submitted on 29 Apr 2019 by Devineni Aslesha
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### Test Suite

Test Status Code Input and Output
1   Pass
s = { ... 'Day Temp' ' 1 1.3' ' 2 1.12' ' 3 17' ' 4 -32' ' 5 13' ' 6 4.4' ' 7 19'}; t_correct = [1.3 1.12 17 -32 13 4.4 19]; assert(isequal(read_and_interp(s),t_correct));

len = 8 s1 = 2×1 char array '' '' s2 = 0 0 0 0 0 0 0 s1 = ' 1.3' s2 = 1.3000 0 0 0 0 0 0 s1 = ' 1.12' s2 = 1.3000 1.1200 0 0 0 0 0 s1 = ' 17' s2 = 1.3000 1.1200 17.0000 0 0 0 0 s1 = ' -32' s2 = 1.3000 1.1200 17.0000 -32.0000 0 0 0 s1 = ' 13' s2 = 1.3000 1.1200 17.0000 -32.0000 13.0000 0 0 s1 = ' 4.4' s2 = 1.3000 1.1200 17.0000 -32.0000 13.0000 4.4000 0 s1 = ' 19' s2 = 1.3000 1.1200 17.0000 -32.0000 13.0000 4.4000 19.0000 t = 1.3000 1.1200 17.0000 -32.0000 13.0000 4.4000 19.0000

2   Pass
s = { ... 'Day Temp' ' 1 1.3' ' 2 1.12' ' 3 17' ' 4 16' ' 5 9999' ' 6 9999' ' 7 19'}; t_correct = [1.3 1.12 17 16 17 18 19]; assert(isequal(read_and_interp(s),t_correct));

len = 8 s1 = 2×1 char array '' '' s2 = 0 0 0 0 0 0 0 s1 = ' 1.3' s2 = 1.3000 0 0 0 0 0 0 s1 = ' 1.12' s2 = 1.3000 1.1200 0 0 0 0 0 s1 = ' 17' s2 = 1.3000 1.1200 17.0000 0 0 0 0 s1 = ' 16' s2 = 1.3000 1.1200 17.0000 16.0000 0 0 0 s1 = ' 9999' s2 = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 9.9990 0 0 s2 = 1.3000 1.1200 17.0000 16.0000 17.0000 0 0 s1 = ' 9999' s2 = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 0.0170 9.9990 0 s2 = 1.3000 1.1200 17.0000 16.0000 17.0000 18.0000 0 s1 = ' 19' s2 = 1.3000 1.1200 17.0000 16.0000 17.0000 18.0000 19.0000 t = 1.3000 1.1200 17.0000 16.0000 17.0000 18.0000 19.0000

3   Pass
s = { ... 'Day Temp' ' 1 -5' ' 2 19' ' 3 1' ' 4 9999' ' 5 3'}; t_correct = [-5 19 1 2 3]; assert(isequal(read_and_interp(s),t_correct));

len = 6 s1 = 2×1 char array '' '' s2 = 0 0 0 0 0 s1 = ' -5' s2 = -5 0 0 0 0 s1 = ' 19' s2 = -5 19 0 0 0 s1 = ' 1' s2 = -5 19 1 0 0 s1 = ' 9999' s2 = -5 19 1 9999 0 s2 = -5 19 1 2 0 s1 = ' 3' s2 = -5 19 1 2 3 t = -5 19 1 2 3