bash 中 for 循环中的简单数学语句
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Simple math statements in bash in a for loop
提问by physicsmichael
I'm quite new to bash scripting and usually avoid it all costs but I need to write a bash script to execute some simple things on a remote cluster. I'm having problems with a for loop that does the following:
我对 bash 脚本很陌生，通常会避免所有成本，但我需要编写一个 bash 脚本来在远程集群上执行一些简单的事情。我在执行以下操作的 for 循环中遇到问题：
for i in {1..20}
do
for j in {1..20}
do
echo (i*i + j*j ) **.5 <--- Pseudo code!
done
done
Can you help me with this simple math? I've thrown $'s everywhere and can't write it properly. If you could help me understand how variables are named/assigned in bash for loops and the limitations of bash math interpretation (how do you do the square root?) I'd be very grateful. Thanks!
你能帮我解决这个简单的数学问题吗？我把$'s 到处扔了，但写不出来。如果您能帮助我了解如何在 bash for 循环中命名/分配变量以及 bash 数学解释的局限性（您如何计算平方根？），我将不胜感激。谢谢！
回答by CB Bailey
Arithmetic expansion needs $((...))notation, so something like:
算术扩展需要$((...))符号，所以像：
echo $((i*i + j*j))
However, bash only uses integers so you may need to use an external tool such as dc.
但是，bash 仅使用整数，因此您可能需要使用外部工具，例如 dc。
E.g.
例如
dc -e "18k $i $i * $j $j * + v p"
回答by dustmachine
Here's a decent solution:
这是一个不错的解决方案：
for i in {1..20}
do
for j in {1..20}
do
echo "scale = 3; sqrt($i*$i + $j*$j)" | bc
done
done
Output will be:
输出将是：
1.414
2.236
3.162
2.236
[...etc...]
回答by ephemient
Shell math can be done in several ways.
壳牌数学可以通过多种方式完成。
echo $(( i*i + j*j ))
echo $[ i*i + j*j ]
expr "$i" '*' "$i" '+' "$j" '*' "$j"
However, this can only handle integer arithmetic. Instead, you can use bc:
但是，这只能处理整数算术。相反，您可以使用bc：
echo "scale = 5; sqrt( $i*$i + $j*$j)" | bc
Change scaleto the number of decimal places desired.
更改scale为所需的小数位数。
回答by DigitalRoss
#!/bin/bash
for i in {1..20}; do
for j in {1..20}; do
echo 5k$i $i\* $j $j\*+vp | dc
done
done
回答by rerun
Use double paren to evaluate a variable.
使用双括号计算变量。
variableA=$((variableB*variableC))
变量A=$((变量B*变量C))
Only for ints though.
仅适用于整数。
回答by ghostdog74
does your remote cluster only have bash? if not, try and see if you have awk
你的远程集群只有 bash 吗？如果没有，请尝试看看您是否有 awk
awk 'BEGIN{
for(i=1;i<=20;i++){
for(j=1;j<=20;j++){
print ( i*i + j*j ) ** 0.5
}
}
}'
回答by Nahuel Fouilleul
with zsh, this will work
使用 zsh，这将起作用
for i in {1..20};do
for j in {1..20};do
echo $((($i*$i + $j*$j)**.5))
done
done
回答by Ben
Typically you would use $((1*3)), but your case won't work as bash doesn't support floating point numbers. You'll have to use an external tool like awk, bc or dc: http://mywiki.wooledge.org/BashFAQ/022
通常您会使用 $((1*3))，但您的情况将不起作用，因为 bash 不支持浮点数。您必须使用外部工具，如 awk、bc 或 dc：http: //mywiki.wooledge.org/BashFAQ/022
回答by bta
The code
编码
echo $[(($i * $i) + ($j * $j)) ** $X]
will work if $Xis an integer. You're trying to take the square root, and I'm not sure if bash's built-in arithmetic will do that. You'll probably be better off using a more powerful calculator tool (like bc, et al.) for this.
如果$X是整数，将起作用。您正在尝试求平方根，我不确定 bash 的内置算术是否会这样做。bc为此，您最好使用功能更强大的计算器工具（如、等）。
回答by wallyk
Bash doesn't offer mathematical functions. However, you almost certainly have the korn shell installed. This should work:
Bash 不提供数学函数。但是，您几乎可以肯定已经安装了 korn shell。这应该有效：
#!/bin/ksh
for i in {1..20}
do
for j in {1..20}
do
x=$((sqrt(i*i + j*j)))
echo "sqrt($i^2 + $j^2) = $x"
done
done
The beginning of the output is
输出的开头是
sqrt(1^2 + 1^2) = 1.41421356237309505
sqrt(1^2 + 2^2) = 2.2360679774997897
sqrt(1^2 + 3^2) = 3.16227766016837933
sqrt(1^2 + 4^2) = 4.12310562561766055
sqrt(1^2 + 5^2) = 5.09901951359278483
sqrt(1^2 + 6^2) = 6.08276253029821969
sqrt(1^2 + 7^2) = 7.07106781186547524
sqrt(1^2 + 8^2) = 8.06225774829854965
sqrt(1^2 + 9^2) = 9.05538513813741663
sqrt(1^2 + 10^2) = 10.0498756211208903
