如果数组元素包含某个字符串，如何从数组中删除这些元素

Question

假设我有一个包含以下数据的数组：

@array[0] = "hello this is a text"
@array[1] = "this is a cat" 
@array[2] = "this is a dog"
@array[3] = "this is a person"
@array[4] = "this is a computer"
@array[5] = "this is a code"
@array[6] = "this is an array"
@array[7] = "this is an element"
@array[8] = "this is a number"

我希望有一个循环，遍历所有数组元素并查找其中是否有值为"dog"的元素。如果存在"dog"，则删除该元素。最终结果如下：

@array[0] = "hello this is a text"
@array[1] = "this is a cat" 
@array[2] = "this is a person"
@array[3] = "this is a computer"
@array[4] = "this is a code"
@array[5] = "this is an array"
@array[6] = "this is an element"
@array[7] = "this is a number"

- dataminer123

---

@var[index] = ... 的风格不佳。请使用 $var[index] = ...。这个表达式使用了 $ 符号，但仍然引用了命名数组 @var。请参见 perldata。 - mob

Answer 1

@array = grep not /dog/, @array;

@array = grep !/dog/, @array;

- mob

---

v5.16.3 给我一个错误提示 "grep 的参数不足"，"not" 应该是一个 "!" 吗？（另一个答案是使用 !） - David Poole

抱歉，David，“not”与“!”的优先级不同。 - mob

Answer 2

显然，重新分配整个数组更容易，但要实际循环删除，可以这样做：

use strict;
use warnings;

my @array = (
    'hello this is a text',
    'this is a cat',
    'this is a dog',
    'this is a person',
    'this is a computer',
    'this is a code',
    'this is an array',
    'this is an element',
    'this is a number'
);

for my $index (reverse 0..$#array) {
    if ( $array[$index] =~ /dog/ ) {
        splice(@array, $index, 1, ());
    }
}

print "$_\n" for @array;

输出：

hello this is a text
this is a cat
this is a person
this is a computer
this is a code
this is an array
this is an element
this is a number

- ysth

---

$#array 是 @array 的最后一个元素的索引（如果为空则为-1）。 - ysth

Answer 3

@array = grep(!/dog/, @array);

- capiggue

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这会同时删除元素吗？ - dataminer123

是的。您可以在此操作之前和之后打印元素（或元素计数）以查看差异。 - capiggue