Python 检查一个列表中的任何元素是否在另一个列表中
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Checking if any elements in one list are in another
提问by h1h1
I'm trying to compare two lists and simply print a message if any value from the first list is in the second list.
我试图比较两个列表，如果第一个列表中的任何值在第二个列表中，则简单地打印一条消息。
def listCompare():
list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]
if list1 in list2:
print("Number was found")
else:
print("Number not in list")
In this example, I want the ifto evaluate to Truebecause 5 is in both lists. This doesn't work, and I'm not sure of the simplest way to compare the two lists.
在此示例中，我希望if计算为True5 ，因为 5 在两个列表中。这不起作用，我不确定比较两个列表的最简单方法。
采纳答案by David Alber
You could solve this many ways. One that is pretty simple to understand is to just use a loop.
你可以通过多种方式解决这个问题。一个非常容易理解的方法是使用循环。
def comp(list1, list2):
for val in list1:
if val in list2:
return True
return False
A more compact way you can do it is to use mapand reduce:
reduce(lambda v1,v2: v1 or v2, map(lambda v: v in list2, list1))
Even better, the reducecan be replaced with any:
更好的是，reduce可以替换为any：
any(map(lambda v: v in list2, list1))
You could also use sets:
您还可以使用集合：
len(set(list1).intersection(list2)) > 0
回答by vidit
There are different ways. If you just want to check if one list contains any element from the other list, you can do this..
有不同的方法。如果您只想检查一个列表是否包含另一个列表中的任何元素，您可以这样做..
not set(list1).isdisjoint(list2)
I believe using isdisjointis better than intersectionfor Python 2.6 and above.
我相信使用isdisjoint比intersectionPython 2.6 及更高版本更好。
回答by dansalmo
Your original approach can work with a list comprehension:
您的原始方法可以与列表理解一起使用：
def listCompare():
list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]
if [item for item in list1 if item in list2]:
print("Number was found")
else:
print("Number not in list")
回答by user1463110
There is a built in function to compare lists:
有一个内置函数来比较列表：
Following is the syntax for cmp() method ?
以下是 cmp() 方法的语法？
cmp(list1, list2)
#!/usr/bin/python
list1, list2 = [123, 'xyz'], [123, 'xyz']
print cmp(list1,list2)
When we run above program, it produces following result ?
当我们运行上面的程序时，它会产生以下结果？
0
If the result is a tie, meaning that 0 is returned
如果结果是平局，则意味着返回 0
回答by colins44
You could change the lists to sets and then compare both sets using the & function. eg:
您可以将列表更改为集合，然后使用 & 函数比较两个集合。例如：
list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]
if set(list1) & set(list2):
print "Number was found"
else:
print "Number not in list"
The "&" operator gives the intersection point between the two sets. If there is an intersection, a set with the intersecting points will be returned. If there is no intersecting points then an empty set will be returned.
“&”运算符给出两个集合之间的交点。如果存在交点，则返回具有交点的集合。如果没有相交点，则将返回一个空集。
When you evaluate an empty set/list/dict/tuple with the "if" operator in Python the boolean False is returned.
当您在 Python 中使用“if”运算符评估空集/列表/字典/元组时，将返回布尔值 False。
回答by Aman Madan
I wrote the following code in one of my projects. It basically compares each individual element of the list. Feel free to use it, if it works for your requirement.
我在我的一个项目中编写了以下代码。它基本上比较列表中的每个单独元素。如果它适合您的要求，请随意使用它。
def reachedGoal(a,b):
if(len(a)!=len(b)):
raise ValueError("Wrong lists provided")
for val1 in range(0,len(a)):
temp1=a[val1]
temp2=b[val1]
for val2 in range(0,len(b)):
if(temp1[val2]!=temp2[val2]):
return False
return True
