如何在python中比较两个有序列表？

If I have one long list: myList = [0,2,1,0,2,1] that I split into two lists:

如果我有一个长列表：myList = [0,2,1,0,2,1]我将其拆分为两个列表：

a = [0,2,1]
b = [0,2,1]

how can I compare these two lists to see if they are both equal/identical, with the constraint that they have to be in the same order?

我如何比较这两个列表，看它们是否相等/相同，约束条件必须是相同的顺序？

I have seen questions asking to compare two lists by sorting them, but in my specific case, I am not checking for a sorted comparison, but identical list comparison.

我已经看到问题要求通过对它们进行排序来比较两个列表，但在我的具体情况下，我没有检查排序比较，但是相同的列表比较。

3 个解决方案

#1

Just use the classic == operator:

只需使用经典的==运算符：

>>> [0,1,2] == [0,1,2]
True
>>> [0,1,2] == [0,2,1]
False
>>> [0,1] == [0,1,2]
False

Lists are equal if elements at the same index are equal. Ordering is taken into account then.

如果同一索引处的元素相等，则列表相等。然后考虑订购。

#2

If you want to just check if they are identical or not, a == b should give you true / false with ordering taken into account.

如果你只想检查它们是否相同，a == b应该给你真/假，并考虑到订购。

In case you want to compare elements, you can use numpy for comparison

如果您想比较元素，可以使用numpy进行比较

c = (numpy.array(a) == numpy.array(b))

c =（numpy.array（a）== numpy.array（b））

Here, c will contain an array with 3 elements all of which are true (for your example). In the event elements of a and b don't match, then the corresponding elements in c will be false.

这里，c将包含一个包含3个元素的数组，所有这些元素都是正确的（对于您的示例）。如果a和b的元素不匹配，则c中的相应元素将为false。

#3

The expression a == b should do the job.

表达式a == b应该完成这项工作。

#1