Question

list = [0, 1, 2, 8, 2, 9, 2]

有没有办法删除元素2，只有一次？

所以你会得到：

list = [0, 1, 2, 8, 9, 2]

我尝试使用index()，但我没有找到它。

它可以是随机2。

所以我无法使用remove()或pop()，因为它不会删除随机位置上的数字2。

Answer 1

这有效

list.remove(2)

L.remove（value） - 删除第一次出现的值。

如果值不存在，则引发ValueError。

Answer 2

使用del或pop

例如，

del list[2]

或

list.pop(2)

del和pop之间的区别在于

del已超载。

例如，del a [1：3]表示删除元素1和3

Answer 3

随机删除<svg version="1.1" id="Layer_1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" x="0px" y="0px" width="841.89px" height="595.28px" viewBox="0 0 841.89 595.28" enable-background="new 0 0 841.89 595.28" xml:space="preserve"> <g> <path class="firstpath" fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" d=" M253.441,284.167c0,0-0.883-14.535,15.429-18.551c5.364-1.32,15.943-0.665,21.667,6.79c13.859,18.051-3.235,32.286-8.087,37.262 c-4.853,4.979-7.713,21.027-17.543,20.528c-9.829-0.497-12.69-6.717-12.317-11.818" /> <path class="secondpath" fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" d=" M285.249,285.252c0,0,0.654-8.212-8.864-10.544c-9.519-2.334-16.796,8.211-10.825,17.449c2.613-1.12,5.226,0.652,5.599,2.52 c0.373,1.866-1.213,4.199-3.826,4.199" /> <path class="thirdpath" fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" d=" M257.815,307.553c0,0-3.732,5.879,2.333,8.118c6.065,2.241,5.318-8.118,8.865-9.237c3.545-1.119,11.943-2.519,11.943-10.265 c0-7.744-6.398-16.451-17.522-11.958" /> </g> <line class="row1col1" fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="319.41" y1="287.987" x2="324.156" y2="286.627" /> <line fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="328.667" y1="285.335" x2="332.744" y2="284.167" /> <line fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="338.25" y1="301.171" x2="343.41" y2="301.171" /> <line fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="328.667" y1="301.171" x2="333.75" y2="301.171" /> <line fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="319.41" y1="301.171" x2="324.156" y2="301.171" /> <line fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="313.741" y1="314.625" x2="319.41" y2="315.987" /> <line fill="none" stroke="#000000" stroke-width="3" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" x1="323.741" y1="316.746" x2="329.667" y2="318.142" /> </svg>

的出现次数

备注：

我们创建了一个数字2的索引列表，即2

使用随机模块choice method从索引列表中随机获取一个索引

列出pop method 或删除方法取决于您的选择

<强>代码：

[i for i, j in enumerate(lst) if j == 2]

<强>输出：

import random lst = [0, 1, 2, 8, 2, 9, 2] lst.pop(random.choice([i for i, j in enumerate(lst) if j == 2])) print lst

Answer 4

请注意，您正在隐藏内置list。除此之外index工作正常：

>>> li = [0, 1, 2, 8, 2, 9, 2]
>>> li.pop(li.index(2))
2
>>> li
[0, 1, 8, 2, 9, 2]

删除列表中的元素

4 个答案: