Cody

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

Solution 1936504

Submitted on 15 Sep 2019 by Julian Lippmann
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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));

temp = 7×1 cell array {'1.3' } {'1.12'} {'17' } {'-32' } {'13' } {'4.4' } {'19' } t = [] i = 1 t = 1.3000 t = 1.3000 1.1200 t = 1.3000 1.1200 17.0000 t = 1.3000 1.1200 17.0000 -32.0000 t = 1.3000 1.1200 17.0000 -32.0000 13.0000 t = 1.3000 1.1200 17.0000 -32.0000 13.0000 4.4000 t = 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));

temp = 7×1 cell array {'1.3' } {'1.12'} {'17' } {'16' } {'9999'} {'9999'} {'19' } t = [] i = 1 t = 1.3000 t = 1.3000 1.1200 t = 1.3000 1.1200 17.0000 t = 1.3000 1.1200 17.0000 16.0000 t = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 9.9990 t = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 9.9990 9.9990 t = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 9.9990 9.9990 0.0190 sep = 1 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));

temp = 5×1 cell array {'-5' } {'19' } {'1' } {'9999'} {'3' } t = [] i = 1 t = -5 t = -5 19 t = -5 19 1 t = -5 19 1 9999 t = -5 19 1 9999 3 sep = 1 t = -5 19 1 2 3