在 Bash 中用 bc 舍入数字
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Rounding Numbers with bc in Bash
提问by Yasin Kaya
I want to compute an average with 3 decimal figures, rounded to nearest, using bc.
我想用 3 位小数计算平均值，四舍五入到最接近，使用bc.
For example:
例如：
average of 3, 3 and 5 should yield 3.667
3、3 和 5 的平均值应为 3.667
and
和
average of 3, 3 and 4 should yield 3.333
3、3 和 4 的平均值应为 3.333
I tried:
我试过：
echo "scale=3; $sum/$n+0.0005" | bc
but scaledoesn't behave as I expect. What can I do to solve my problem?
但scale不符合我的预期。我能做些什么来解决我的问题？
采纳答案by gniourf_gniourf
Your trick to add 0.0005is not a bad idea. Though, it doesn't quite work that way. scaleis used internally when bcperforms some operations (like divisions).
你添加的技巧0.0005不是一个坏主意。尽管如此，它并不完全如此。scale在bc执行某些操作（如除法）时在内部使用。
In your case, it would be better to perform the division first, maybe using a large scaleor the -lswitch to bc1(if your version supports it), then add 0.0005and then set scale=3and perform an operation involving scaleinternally to have the truncation performed.
在您的情况下，最好先执行除法，可能使用大scale或-l切换到bc1（如果您的版本支持），然后添加0.0005然后设置scale=3并执行涉及scale内部的操作以执行截断。
Something like:
就像是：
`a=$sum/$n+0.0005; scale=3; a/1`
Of course, you'll want to proceed differently whether sumis positive or negative. Fortunately, bchas some conditional operators.
当然，无论sum是积极的还是消极的，您都希望以不同的方式进行。幸运的是，bc有一些条件运算符。
`a=$sum/$n; if(a>0) a+=0.0005 else if (a<0) a-=0.0005; scale=3; a/1`
You'll then want to format this answer using printf.
然后，您需要使用printf.
Wrapped in a function round(where you can optionally select the number of decimal figures):
封装在一个函数中round（您可以在其中选择小数位数）：
round() {
# is expression to round (should be a valid bc expression)
# is number of decimal figures (optional). Defaults to three if none given
local df=${2:-3}
printf '%.*f\n' "$df" "$(bc -l <<< "a=; if(a>0) a+=5/10^($df+1) else if (a<0) a-=5/10^($df+1); scale=$df; a/1")"
}
Try it:
尝试一下：
gniourf$ round "(3+3+4)/3"
3.333
gniourf$ round "(3+3+5)/3"
3.667
gniourf$ round "-(3+3+5)/3"
-3.667
gniourf$ round 0
0.000
gniourf$ round 1/3 10
0.3333333333
gniourf$ round 0.0005
0.001
gniourf$ round 0.00049
0.000
1with the -lswitch, scaleis set to 20, which should be plenty enough.
1与-l开关，scale设置为20，应该足够了。
回答by Serge3leo
Next function round argument 'x' to 'd' digits:
下一个函数将参数 'x' 转换为 'd' 数字：
define r(x, d) {
auto r, s
if(0 > x) {
return -r(-x, d)
}
r = x + 0.5*10^-d
s = scale
scale = d
r = r*10/10
scale = s
return r
}
回答by poinck
This solution is not that flexibile (it just converts float to int), but it can deal with negative numbers:
这个解决方案不是那么灵活（它只是将 float 转换为 int），但它可以处理负数：
e=$( echo "scale=0; (${e}+0.5)/1" | bc -l )
if [[ "${e}" -lt 0 ]] ; then
e=$(( e - 1 ))
fi
