Linux 在 Bash 中舍入除数

Round a divided number in Bash

提问by Mint

How would I round the result from two divided numbers, e.g.

我将如何从两个相除的数字中取整结果，例如

3/2

As when I do

就像我做的那样

testOne=$((3/2))

$testOne contains "1" when it should have rounded up to "2" as the answer from 3/2=1.5

$testOne 包含“1”，而它应该四舍五入为“2”作为 3/2=1.5 的答案

采纳答案by Ben Voigt

To do rounding up in truncating arithmetic, simply add (denom-1)to the numerator.

要在截断算术中进行四舍五入，只需添加(denom-1)到分子。

Example, rounding down:

示例，四舍五入：

N/2
M/5
K/16

Example, rounding up:

示例，四舍五入：

(N+1)/2
(M+4)/5
(K+15)/16

To do round-to-nearest, add (denom/2)to the numerator (halves will round up):

要进行四舍五入，请添加(denom/2)到分子（一半将四舍五入）：

(N+1)/2
(M+2)/5
(K+8)/16

回答by ghostdog74

bash will not give you correct result of 3/2 since it doesn't do floating pt maths. you can use tools like awk

bash 不会给你 3/2 的正确结果，因为它不做浮动 pt 数学。你可以使用awk之类的工具

$ awk  'BEGIN { rounded = sprintf("%.0f", 3/2); print rounded }'
2

or bc

或公元前

$ printf "%.0f" $(echo "scale=2;3/2" | bc)
2

回答by Sean Staats

Another solution is to do the division within a python command. For example:

另一种解决方案是在 python 命令中进行除法。例如：

$ numerator=90
$ denominator=7
$ python -c "print (round(${numerator}.0 / ${denominator}.0))"

Seems less archaic to me than using awk.

对我来说似乎不如使用 awk 过时。

回答by Ashish

Good Solution is to get Nearest Round Number is

好的解决方案是获得最近的整数是

var=2.5
echo $var | awk '{print int(+0.5)}'

Logic is simple if the var decimal value is less then .5 then closest value taken is integer value. Well if decimal value is more than .5 then next integer value gets added and since awk then takes only integer part. Issue solved

如果 var 十进制值小于 0.5，则逻辑很简单，则采用的最接近的值是整数值。好吧，如果十进制值大于 0.5，则添加下一个整数值，因为 awk 只取整数部分。问题已解决

回答by guest

I think this should be enough.

我认为这应该足够了。

$ echo "3/2" | bc 

回答by Kaz

If you have integer division of positive numbers which rounds toward zero, then you can add one less than the divisor to the dividend to make it round up.

如果您有向零舍入的正数的整数除法，那么您可以在被除数上加上比除数少 1 以使其四舍五入。

That is to say, replace X / Ywith (X + Y - 1) / Y.

也就是说，替换X / Y为(X + Y - 1) / Y。

Proof:

证明：

• Case 1: X = k * Y(X is integer multiple of Y): In this case, we have (k * Y + Y - 1) / Y, which splits into (k * Y) / Y + (Y - 1) / Y. The (Y - 1)/Ypart rounds to zero, and we are left with a quotient of k. This is exactly what we want: when the inputs are divisible, we want the adjusted calculation to still produce the correct exact quotient.

• Case 2: X = k * Y + mwhere 0 < m < Y(X is not a multiple of Y). In this case we have a numerator of k * Y + m + Y - 1, or k * Y + Y + m - 1, and we can write the division out as (k * Y)/Y + Y/Y + (m - 1)/Y. Since 0 < m < Y, 0 <= m - 1 < Y - 1, and so the last term (m - 1)/Ygoes to zero. We are left with (k * Y)/Y + Y/Ywhich work out to k + 1. This shows that the behavior rounds up. If we have an Xwhich is a kmultiple of Y, if we add just 1 to it, the division rounds up to k + 1.

• 情况 1：X = k * Y（X 是 Y 的整数倍）：在这种情况下，我们有(k * Y + Y - 1) / Y，它分成(k * Y) / Y + (Y - 1) / Y。该(Y - 1)/Y部分四舍五入为零，我们剩下的商为k。 这正是我们想要的：当输入可整除时，我们希望调整后的计算仍然产生正确的精确商。

• 情况 2：X = k * Y + m其中0 < m < Y（X 不是 Y 的倍数）。在这种情况下，我们的分子为k * Y + m + Y - 1, 或k * Y + Y + m - 1，我们可以将除法写为(k * Y)/Y + Y/Y + (m - 1)/Y。因为0 < m < Y, 0 <= m - 1 < Y - 1, 所以最后一项(m - 1)/Y变为零。我们剩下的(k * Y)/Y + Y/Y工作是k + 1。这表明该行为四舍五入。如果我们有一个X是 的k倍数Y，如果我们只给它加 1，则除法四舍五入为k + 1。

But this rounding is extremely opposite; all inexact divisions go away from zero. How about something in between?

但这种四舍五入是极其相反的；所有不精确的除法都从零开始。介于两者之间怎么样？

That can be achieved by "priming" the numerator with Y/2. Instead of X/Y, calculate (X+Y/2)/Y. Instead of proof, let's go empirical on this one:

这可以通过“启动”分子来实现Y/2。而不是X/Y，计算(X+Y/2)/Y。代替证明，让我们对这一点进行实证：

$ round()
> {
>   echo $((( + /2) / ))
> }
$ round 4 10
0
$ round 5 10
1
$ round 6 10
1
$ round 9 10
1
$ round 10 10
1
$ round 14 10
1
$ round 15 10
2

Whenever the divisor is an even, positive number, if the numerator is congruent to half that number, it rounds up, and rounds down if it is one less than that.

当除数是偶数、正数时，如果分子等于该数的一半，则向上舍入，如果小于 1，则向下舍入。

For instance, round 6 12goes to 1, as do all values which are equal to 6, modulo 12, like 18(which goes to 2) and so on. round 5 12goes down to 0.

例如，round 6 12转到1，所有等于6，取模的值12，如18（转到 2）等等。 round 5 12归结为0.

For odd numbers, the behavior is correct. None of the exact rational numbers are midway between two consecutive multiples. For instance, with a denominator of 11we have 5/11 < 5.5/11 (exact middle) < 6/11; and round 5 11rounds down, whereas round 6 11rounds up.

对于奇数，行为是正确的。没有一个精确的有理数介于两个连续的倍数之间。例如，分母为11我们有5/11 < 5.5/11 (exact middle) < 6/11；并向下round 5 11舍入，而round 6 11向上舍入。

回答by Q-D

If the decimal separator is comma (eg : LC_NUMERIC=fr_FR.UTF-8, see here):

如果小数点分隔符是逗号（例如：LC_NUMERIC=fr_FR.UTF-8，请参见此处）：

$ printf "%.0f" $(echo "scale=2;3/2" | bc) 
bash: printf: 1.50: nombre non valable
0

Substitution is needed for ghostdog74 solution :

ghostdog74 解决方案需要替换：

$ printf "%.0f" $(echo "scale=2;3/2" | bc | sed 's/[.]/,/')
2

or

或者

$ printf "%.0f" $(echo "scale=2;3/2" | bc | tr '.' ',')
2

回答by sk8asd123

To round up you can use modulus.

要四舍五入，您可以使用模数。

The second part of the equation will add to True if there's a remainder. (True = 1; False = 0)

如果有余数，等式的第二部分将与 True 相加。（真 = 1；假 = 0）

ex: 3/2

例如：3/2

answer=$(((3 / 2) + (3 % 2 > 0)))
echo $answer
2

ex: 100 / 2

例如：100 / 2

answer=$(((100 / 2) + (100 % 2 > 0)))
echo $answer
50

ex: 100 / 3

例如：100 / 3

answer=$(((100 / 3) + (100 % 3 > 0)))
echo $answer
34

回答by N. Ingersoll

Given a floating point value, we can round it trivially with printf:

给定一个浮点值，我们可以用 printf 对其进行四舍五入：

# round  to  decimal places
round() {
    printf "%.{:-0}f" ""
}

Then,

然后，

# do some math, bc style
math() {
    echo "$*" | bc -l
}

$ echo "Pi, to five decimal places, is $(round $(math "4*a(1)") 5)"
Pi, to five decimal places, is 3.14159

Or, to use the original request:

或者，使用原始请求：

$ echo "3/2, rounded to the nearest integer, is $(round $(math "3/2") 0)"
3/2, rounded to the nearest integer, is 2

回答by xs2rashid

Following worked for me.

以下为我工作。

 #!/bin/bash
function float() {
bc << EOF
num = ;
base = num / 1;
if (((num - base) * 10) > 1 )
    base += 1;
print base;
EOF
echo ""
}

float 3.2