bash 用bash中的数字序列创建一个数组
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Create an array with a sequence of numbers in bash
提问by jarhead
I would like to write a script that will create me an array with the following values:
我想编写一个脚本，该脚本将为我创建一个具有以下值的数组：
{0.1 0.2 0.3 ... 2.5}
Until now I was using a script as follows:
到目前为止，我使用的脚本如下：
plist=(0.1 0.2 0.3 0.4)
for i in ${plist[@]}; do
echo "submit a simulation with this parameter:"
echo "$i"
done
But now I need the list to be much longer ( but still with constant intervals).
但是现在我需要更长的列表（但仍然具有恒定的间隔）。
Is there a way to create such an array in a single command? what is the most efficient way to create such a list?
有没有办法在单个命令中创建这样的数组？创建此类列表的最有效方法是什么？
回答by fedorqui 'SO stop harming'
Using seqyou can say seq FIRST STEP LAST. In your case:
使用seq你可以说seq FIRST STEP LAST。在你的情况下：
seq 0 0.1 2.5
Then it is a matter of storing these values in an array:
然后是将这些值存储在数组中的问题：
vals=($(seq 0 0.1 2.5))
You can then check the values with:
然后，您可以使用以下方法检查值：
$ printf "%s\n" "${vals[@]}"
0,0
0,1
0,2
...
2,3
2,4
2,5
Yes, my locale is set to have commas instead of dots for decimals. This can be changed setting LC_NUMERIC="en_US.UTF-8".
是的，我的语言环境设置为逗号而不是小数点。这可以更改设置LC_NUMERIC="en_US.UTF-8"。
By the way, brace expansionalso allows to set an increment. The problem is that it has to be an integer:
顺便说一下，大括号扩展也允许设置增量。问题是它必须是一个整数：
$ echo {0..15..3}
0 3 6 9 12 15
回答by Amir Mehler
Bash supports C style For loops:
Bash 支持 C 风格的 For 循环：
$ for ((i=1;i<5;i+=1)); do echo "0.${i}" ; done
0.1
0.2
0.3
0.4
回答by ePi272314
Complementing the main answer
补充主要答案
In my case, seqwas not the best choice.
To produce a sequence, you can also use the jotutility. However, this command has a more elaborated syntaxis.
就我而言，seq这不是最佳选择。
要生成序列，您还可以使用该jot实用程序。但是，此命令具有更详细的语法。
# 1 2 3 4
jot - 1 4
# 3 evenly distributed numbers between 0 and 10
# 0 5 10
jot 3 0 10
# a b c ... z
jot -c - 97 122
