每个字符串每出现第 2 次，分别用文件的第 n 行替换两个不同的字符串

Posted 2023-03-24

【中文标题】每个字符串每出现第 2 次，分别用文件的第 n 行替换两个不同的字符串

【英文标题】：replace two different strings with the nth line of a file every 2 nth occurrences of each string separately

【发布时间】：2021-12-20 08:17:45

【问题描述】：

我正在尝试使用以下command_line01 将home_cool 的第一次和第二次出现分别替换为1.txt 的第一行，然后继续替换home_cool01 的第一次和第二次出现也分别由1.txt 的第一行，然后..，用1.txt 的第二行单独替换第三和第四次出现的home_cool。等等，即每2 nth 出现home_cool 或home_cool01，分别用1.txt 的第n 行替换两个字符串。

我试过command_line01 波纹管：

awk 'NR==FNR a[NR]=$0; next /home_cool01/gsub("home_cool01", a[++i<3]) /home_cool/gsub("home_cool", a[++j<3]) 1' 1.txt 0.txt > 2.txt

但这仅适用于 前两次出现，因为接下来的两次出现 home_cool 或 home_cool01 被 nothing 替换，如下所示：

"#sun\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree()\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree()\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree()\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree()\t",
"machine(shoes_shirt.shop)\t",

这是我的两个源文件：

0.txt：

"#sun\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tre(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",

和

1.txt：

(food, apple, sky, cat,blue,)(bag, tortoise,)
(food, apple, sky, cat,blue,)(bag,)
(food, apple, sky, cat,blue,)(bag, moon, tortoise,)

而我想要的输出 2.txt 是：

"#sun\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",

编辑：来自@markp-fuso cmets：

1 - 如果有多个条目，如果已​​经完成第一个条目的处理，我将从第二个重新启动。

2 - 如果我有超过 6 个 home_coool 条目...我从 1.txt 的开头继续。

3 - 我不想仅限于两种研究模式，所以当有home_coool02、home_coool03、...、home_coool_some_sufix 时，它应该是一个合适的解决方案，但我需要保持我最初发布的 MWE

【问题讨论】：

如果输入是home_cool、home_cool01、home_cool 怎么办...您是重新启动第二个home_cool 的计数器还是从上次中断的地方继续？如果您有超过 6 个 home_cool 条目会发生什么...您是继续使用来自 1.txt 的最后一个条目还是从 1.txt 的开头重新开始？您是否需要担心其他搜索模式，例如home_cool02、home_cool04、home_cool_some_other_suffix 等？

我认为您可能需要阅读有关 % 运算符的信息。

@Luuk 我会阅读谢谢！

@markp-fuso 对您的第一条评论： 1 - 如果有多个条目，如果已​​经完成第一个条目的处理，我将从第二个条目重新启动。 2 - 如果我有超过 6 个 home_coool 条目......我从 1.txt 的开头继续。 3 - 我不想仅限于两种研究模式，所以当有home_coool02，home_coool03，...，home_coool_some_sufix时，它应该是一个合适的解决方案，但我需要保持一个MWE作为我最初发布了

【参考方案1】：

假设：

只需要担心搜索模式 home_cool 和 home_cool01（可以添加更多，但需要一些返工；可能使用关联数组来跟踪每个唯一模式的计数） 在继续下一个替换模式之前，将应用两次替换模式 如果我们到达替换模式的末尾，我们会从头开始

示例输入：

$ cat 0.txt
"#sun\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool01)\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",          # pick up where we left off with 'home_cool'
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",          # restart from beginning of 0.txt replacment patterns
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree(home_cool)\t",
"machine(shoes_shirt.shop)\t",

一个awk想法：

awk '
NR==FNR        a[NR]=$0; n=NR; next
/home_cool01/  gsub("home_cool01", a[int((i++)%(n*2)/2)+1])
/home_cool/    gsub("home_cool",   a[int((j++)%(n*2)/2)+1])
1
' 1.txt 0.txt

地点：

n==3 - 1.txt 中的行数；所以n*2 == 6 %6 - 将生成 0-5 的输出 (%6/2) - 将输出转换为 0.0, 0.5, 1.0, 1.5, 2.0, 2.5 int(...) - 将输出变成0, 0, 1, 1, 2, 2 +1 - 为我们提供1, 1, 2, 2, 3, 3 的数组索引

注意：

是的，索引有点复杂；我愿意接受简化的建议

这会生成：

"#sun\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, moon, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, moon, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",
"car_snif = house.group_tree((food, apple, sky, cat,blue,)(bag, tortoise,))\t",
"machine(shoes_shirt.shop)\t",

【讨论】：

我正在做测试，现在效果很好，感谢您的解释

你的解释帮助我理解了其他 UNIX 操作，很好的解释..

我有一个新需求：假设不是用 nth 的 1.txt 文件行替换 0.txt 文件模式中的 nth's，现在我想要 0.txt 和 @ 之间的过程987654342@ 只发生在1.txt 的第二行用作输出之前，此时我将进程重定向到另一个包含要替换的home_cool 字符串的文件。我应该为此发布一个新问题吗？

是的，听起来像是一个新问题，特别是因为我（还）不明白你的提议......所以我假设会有更多轮次的来回......这只会使这个特定的问答变得混乱；是的，如果您打算将另一个文件加入其中，添加逻辑以在替换文件之间跳转……绝对是一个全新的问答

@mark-fuso 是的，我需要用更好的方式表达这一点，我将花一些时间来处理这个问题。