Question

如果数字超过阈值（不是0.5）并且向下舍入，我希望能够“舍入”一个数字。

这是我提出的一些糟糕的代码。在matlab中是否有内置函数，或更优雅的解决方案（矢量化可能）？

function [ rounded_numbers ] = custom_round( input_numbers, threshold )
%CUSTOM_ROUND rounds between 0 and 1 with threshold threshold

  [input_rows, input_cols] = size(input_numbers);
  rounded_numbers = zeros(input_rows, input_cols);

  for i = 1:length(input_numbers)
    if input_numbers(i) > threshold
      rounded_numbers(i) = 1;
    else
      rounded_numbers(i) = 0;
    end
  end
end

由于

Answer 1

只需使用

round(x-treshold+0.5)

测试用例：

>> x = -10:0.3:10
ans =
    -2   -1.7  -1.4  -1.1  -0.8  -0.5  -0.2  0.1    0.4   0.7    1    1.3   1.6   1.9


>> treshold = 0.8; % round everything up for which holds mod(x,1) >= treshold
>> y = round(x-treshold+0.5)

ans =
    -2    -2    -2    -1    -1    -1    -1     0     0     0     1     1     1     2

负数也正确舍入，边界除外：-0.8舍入为-1而不是0，但这与圆正常的行为相同：round（-0.5）返回-1

Answer 2

这是一个解决方案，如果数字已超过阈值

，我们将从零舍入

in = [0.2,-3.3,4.1];
th = 0.2;

%# get the fractional part of the number
frac = mod(in,1); %# positive for negative in

%# find the sign so we know whether to round
%# to plus or minus inf
sig = sign(in);

%# identify which way to round
upIdx = frac>th; %# at threshold, we round down

%# round towards inf if up
out = abs(in);
out(upIdx) = ceil(out(upIdx));
out(~upIdx) = floor(out(~upIdx));
%# re-set the sign
out= out.*sig
out =
 0    -4     4

注意：如果数字仅在0和1之间，则更容易：

%# this does exactly what your code does
out = double(in>th);

Answer 3

这适用于任何数字，而不仅仅是0到1之间。阈值必须在[0,1]范围内。

我没有测试过负数。

function [result] = custom_round( num, threshold )

if ( threshold < 0 ) || ( threshold >= 1 )
  error( 'threshold input must be in the range [0,1)' );
end

fractional = num - floor( num );
idx1 = fractional > threshold;
idx2 = fractional <= threshold;
difference = 1 - fractional;
result = num + ( difference .* idx1 ) - ( fractional .* idx2 );

end

测试

>> custom_round( [0.25 0.5 0.75 1], 0.3 )
ans =
     0     1     1     1

>> custom_round( [0.25 0.5 0.75 1], 0.8 )
ans =
     0     0     0     1

>> custom_round( [10.25 10.5 10.75 11], 0.8 )
ans =
    10    10    10    11

使用自定义阈值舍入数字

3 个答案: