Rounding decimals matlab manipulates
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I need to write, in 1 code line, given a number (pos, neg, int, float), how do I round it without preset functions like fix, round, etc.?
9 件のコメント
Walter Roberson
2017 年 7 月 31 日
編集済み: Walter Roberson
2017 年 7 月 31 日
How would you do it in multiple lines of code?
hint 1: watch out for negative
hint 2: log10 to determine the number of decimal places it already has
hint 3: watch out for 0
hint 4: watch out for integers that are exact powers of 10: for example, log10(100) = 2 exactly, log10(99.999) = slightly less than 2.
hint 5: x * (1-eps) for positive x is always less than x
Stephen23
2017 年 8 月 1 日
Well thank for the hints Accualy in one code line zerp loops. Smone can help?
Guillaume
2017 年 8 月 1 日
@Yuval,
a) Use Comment on this question rather than starting an answer
b) proofread what you read. The above is incomprehensible.
Walter Roberson
2017 年 8 月 1 日
Is floor() permitted?
yuval ohayon
2017 年 8 月 1 日
yuval ohayon
2017 年 8 月 1 日
Steven Lord
2017 年 8 月 2 日
In general people on MATLAB Answers don't post answers to homework questions or questions that sound like homework. If you show what you've done to try to solve the problem on your own and ask a specific question about where you're having difficulty you may receive some more specific suggestions.
Of course, if this isn't a homework problem, just use the round function or the other rounding functions included in MATLAB.
Image Analyst
2017 年 8 月 2 日
mod() is also a built in function like round(), fix(), int32(), ceil(), floor(), etc.
回答 (1 件)
Jan
2017 年 8 月 1 日
Try it:
x = 13.3
y = mod(x, 1)
z = 13.6
y = mod(x, 1)
Now you should see how to use the output to decide, if the rest of the division by 1 is lower or higher than 0.5.
3 件のコメント
yuval ohayon
2017 年 8 月 1 日
Walter Roberson
2017 年 8 月 1 日
x - mod(x,1) + mod(x,1)>0.5
Jan
2017 年 8 月 2 日
int64() rounds also. And you can use cast() to preserve the original class.
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