从 linux (BASH) 中的数字获取上限整数
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Get ceiling integer from number in linux (BASH)
提问by Mint
How would I do something like:
我将如何做类似的事情：
ceiling(N/500)
N representing a number.
N 代表一个数字。
But in a linux Bash script
但是在 linux Bash 脚本中
采纳答案by Josh McFadden
Call out to a scripting language with a ceil function. Given $NUMBER:
使用 ceil 函数调用脚本语言。鉴于$NUMBER：
python -c "from math import ceil; print ceil($NUMBER/500.0)"
or
或者
perl -w -e "use POSIX; print ceil($NUMBER/500.0), qq{\n}"
回答by Kalle
Why use external script languages? You get floor by default. To get ceil, do
为什么要使用外部脚本语言？默认情况下，您获得发言权。要获得细胞，请执行
$ divide=8; by=3; let result=($divide+$by-1)/$by; echo $result
3
$ divide=9; by=3; let result=($divide+$by-1)/$by; echo $result
3
$ divide=10; by=3; let result=($divide+$by-1)/$by; echo $result
4
$ divide=11; by=3; let result=($divide+$by-1)/$by; echo $result
4
$ divide=12; by=3; let result=($divide+$by-1)/$by; echo $result
4
$ divide=13; by=3; let result=($divide+$by-1)/$by; echo $result
5
....
To take negative numbers into account you can beef it up a bit. Probably cleaner ways out there but for starters
要考虑负数，您可以稍微加强一下。可能更清洁的方法，但对于初学者来说
$ divide=-10; by=10; neg=; if [ $divide -lt 0 ]; then let divide=-$divide; neg=1; fi; let result=($divide+$by-1)/$by; if [ $neg ]; then let result=-$result; fi; echo $result
-1
$ divide=10; by=10; neg=; if [ $divide -lt 0 ]; then let divide=-$divide; neg=1; fi; let result=($divide+$by-1)/$by; if [ $neg ]; then let result=-$result; fi; echo $result
1
回答by Harvey
Here's a solution using bc (which should be installed just about everywhere):
这是一个使用 bc 的解决方案（它应该安装在几乎所有地方）：
ceiling_divide() {
ceiling_result=`echo "( + - 1)/" | bc`
}
Here's another purely in bash:
这是另一个纯粹的 bash：
# Call it with two numbers.
# It has no error checking.
# It places the result in a global since return() will sometimes truncate at 255.
# Short form from comments (thanks: Jonathan Leffler)
ceiling_divide() {
ceiling_result=$(((+-1)/))
}
# Long drawn out form.
ceiling_divide() {
# Normal integer divide.
ceiling_result=$((/))
# If there is any remainder...
if [ $((%)) -gt 0 ]; then
# rount up to the next integer
ceiling_result=$((ceiling_result + 1))
fi
# debugging
# echo $ceiling_result
}
回答by ghostdog74
You can use awk
你可以使用 awk
#!/bin/bash
number=""
divisor=""
ceiling() {
awk -vnumber="$number" -vdiv="$divisor" '
function ceiling(x){return x%1 ? int(x)+1 : x}
BEGIN{ print ceiling(number/div) }'
}
ceiling
output
输出
$ ./shell.sh 1.234 500
1
Or if there's a choice, you can use a better shell that does floating point, eg Zsh
或者，如果有选择，您可以使用更好的浮点外壳，例如 Zsh
integer ceiling_result
ceiling_divide() {
ceiling_result=$((/))
echo $((ceiling_result+1))
}
ceiling_divide 1.234 500
回答by Frank R.
Mathematically, the function of ceiling can be define with floor, ceiling(x) = -floor(-x). And, floor is the default when converting a positive float to integer.
在数学上，天花板的函数可以用地板来定义，天花板(x) = -地板(-x)。而且，floor 是将正浮点数转换为整数时的默认值。
if [ $N -gt 0 ]; then expr 1 - $(expr $(expr 1 - $N) / 500); else expr $N / 500; fi
Ref. https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
参考 https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
回答by bsaldivar
Floor () {
DIVIDEND=
DIVISOR=
RESULT=$(( ( ${DIVIDEND} - ( ${DIVIDEND} % ${DIVISOR}) )/${DIVISOR} ))
echo ${RESULT}
}
R=$( Floor 8 3 )
echo ${R}
Ceiling () {
DIVIDEND=
DIVISOR=
$(( ( ( ${DIVIDEND} - ( ${DIVIDEND} % ${DIVISOR}) )/${DIVISOR} ) + 1 ))
echo ${RESULT}
}
R=$( Ceiling 8 3 )
echo ${R}
回答by MestreLion
Expanding a bit on Kalle's great answer, here's the algorithm nicely packed in a function:
扩展一下Kalle 的好答案，这是一个很好地打包在一个函数中的算法：
ceildiv() {
local num=
local div=
echo $(( (num + div - 1) / div ))
}
or as a one-liner:
或作为单线：
ceildiv(){ echo $(((+-1)/)); }
If you want to get fancy, you could use a more robust version validates input to check if they're numerical, also handles negative numbers:
如果你想花哨，你可以使用更强大的版本验证输入来检查它们是否是数字，也可以处理负数：
ceildiv() {
local num=${1:-0}
local div=${2:-1}
if ! ((div)); then
return 1
fi
if ((num >= 0)); then
echo $(( (num + div - 1) / div ))
else
echo $(( -(-num + div - 1) / div ))
fi
}
This uses a "fake" ceil for negative numbers, to the highest absoluteinteger, ie, -10 / 3 = -4 and not -3 as it should, as -3 > -4. If you want a "true" ceil, use $(( num / div ))instead after the else
这对负数使用“假”ceil 到最高绝对整数，即 -10 / 3 = -4 而不是 -3，因为 -3 > -4。如果你想要一个“真正的”ceil，请$(( num / div ))在else
And then use it like:
然后像这样使用它：
$ ceildiv 10 3 4 $ ceildiv 501 500 2 $ ceildiv 0 3 0 $ ceildiv -10 1 -10 $ ceildiv -10 3 -4
回答by splaisan
Using the gorgeous 'printf' 1will round up to the next integer
使用华丽的 'printf' 1将四舍五入到下一个整数
printf %.0f $float
or
printf %.0f `your calculation formula`
or
printf %.0f $(your calculation formula)
ref: how to remove decimal from a variable?
ref：如何从变量中删除小数？
回答by Kemin Zhou
Without specifying any function, we can use the following awk script:
在不指定任何函数的情况下，我们可以使用以下 awk 脚本：
echo x y | awk '{ r= % ; q=/y; if (r != 0) q=int(q+1); print q}'
Not sure this one get any logical error or not. Please correct.
不确定这个是否有任何逻辑错误。请纠正。
回答by Leo
If you are already familiar with the Python library, then rather than learn bc, you might want to define this bash function:
如果您已经熟悉 Python 库，那么bc您可能想要定义这个 bash 函数而不是学习：
pc () { pyexpr="from math import *; print($@)"; python -c "$pyexpr"; }
Then:
然后：
pc "ceil(3/4)"
1
but also any valid python expressionworks:
但任何有效的 python表达式也有效：
pc pi / 4
0.7853981633974483
pc "'\n'.join(['Pythagoras said that %3.2f^2 + %3.2f^2 is always %3.2f'
% (sin(ai), cos(ai), sin(ai)**2 + cos(ai)**2)
for ai in [pi / 4 * k for k in range(8)]])"
Pythagoras said that 0.00^2 + 1.00^2 is always 1.00
Pythagoras said that 0.71^2 + 0.71^2 is always 1.00
Pythagoras said that 1.00^2 + 0.00^2 is always 1.00
Pythagoras said that 0.71^2 + -0.71^2 is always 1.00
Pythagoras said that 0.00^2 + -1.00^2 is always 1.00
Pythagoras said that -0.71^2 + -0.71^2 is always 1.00
Pythagoras said that -1.00^2 + -0.00^2 is always 1.00
Pythagoras said that -0.71^2 + 0.71^2 is always 1.00
