Cody

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

Solution 2247543

Submitted on 1 May 2020 by shaikh akbar shaikh rasul
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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 = ' 1 1.3' temp1 = 1.3000 t = 1.3000 temp = ' 2 1.12' temp1 = 1.1200 t = 1.3000 1.1200 temp = ' 3 17' temp1 = 17 t = 1.3000 1.1200 17.0000 temp = ' 4 -32' temp1 = -32 t = 1.3000 1.1200 17.0000 -32.0000 temp = ' 5 13' temp1 = 13 t = 1.3000 1.1200 17.0000 -32.0000 13.0000 temp = ' 6 4.4' temp1 = 4.4000 t = 1.3000 1.1200 17.0000 -32.0000 13.0000 4.4000 temp = ' 7 19' temp1 = 19 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 = ' 1 1.3' temp1 = 1.3000 t = 1.3000 temp = ' 2 1.12' temp1 = 1.1200 t = 1.3000 1.1200 temp = ' 3 17' temp1 = 17 t = 1.3000 1.1200 17.0000 temp = ' 4 16' temp1 = 16 t = 1.3000 1.1200 17.0000 16.0000 temp = ' 5 9999' temp1 = 9999 t = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 9.9990 t = 1.3000 1.1200 17.0000 16.0000 17.0000 temp = ' 6 9999' temp1 = 9999 t = 1.0e+03 * 0.0013 0.0011 0.0170 0.0160 0.0170 9.9990 t = 1.3000 1.1200 17.0000 16.0000 17.0000 18.0000 temp = ' 7 19' temp1 = 19 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 = ' 1 -5' temp1 = -5 t = -5 temp = ' 2 19' temp1 = 19 t = -5 19 temp = ' 3 1' temp1 = 1 t = -5 19 1 temp = ' 4 9999' temp1 = 9999 t = -5 19 1 9999 t = -5 19 1 2 temp = ' 5 3' temp1 = 3 t = -5 19 1 2 3

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