有没有一种方法可以在Python中跳过itertools.product循环中的值？

我正在制作一个脚本来比较两个地址之间的距离，这些地址与

itertools.product

一起存储在两个不同的.csv文件中

比较是一种检查地址之间是否相距1km的

if

条件

需要将

地址1

中的每个地址与

地址2

中的每个地址进行比较

如果

addresses1

中的地址与

addresses2

中的地址相距1km，则需要移动到下一个地址。由于列表太长，等待移动到下一个地址是浪费时间的。-这就是问题所在

我尝试了

继续

，但不起作用

假设我有两个地址列表：

addresses1 = ["Address1","Address2","Address3","Address4"]
addresses2 = ["compAddress1","compAddress2,"compAddress3","compAddress4","compAddress5"]

producted_list = list(itertools.product(addresses1, addresses2))

for a,b in producted_list:
    BLOCK OF CODE WITH GEOLOCATIONS

    if(distance == 1km):
        print("Addresses are within 1km from each other, move on next one")
        continue

让我们看看它是如何比较的：

a[0] -> b[0]
a[0] -> b[1]
a[0] -> b[2]
a[0] -> b[3]
a[0] -> b[4]

a[1] -> b[0]
a[1] -> b[1]
a[1] -> b[2]
a[1] -> b[3]
a[1] -> b[4]

a[2] -> b[0]
a[2] -> b[1]
a[2] -> b[2]
a[2] -> b[3]
a[2] -> b[4]
.
.
.

假设

a[1]

与

b[0]

匹配，我想立即继续

a[2]

，而不必继续

a[1]

与

b[]

我试着使用Eric的代码

与

对于b中的a：
def doWork（）：
对于d中的c：
如果出现以下情况：
返回#
嫁妆

---

但它不工作，它会中断，或者它需要更改。

只需写两个循环，然后从内部循环中断以继续外部循环：

addresses1 = ["Address1", "Address2", "Address3", "Address4"]
addresses2 = ["compAddress1", "compAddress2", "compAddress3", "compAddress4", "compAddress5"]

for a in addresses1:
    for b in addresses2:
        BLOCK OF CODE WITH GEOLOCATIONS

        if(distance == 1km):
            print("Addresses are within 1km from each other, move on next one")
            break

for a in addresses_1:
    for b in addresses_2:
        if distance(a, b) <= 1.00:
            print(a, b, "are within 1 km")
            break

---

只需编写两个循环，并中断内部循环，以继续外部循环：

addresses1 = ["Address1", "Address2", "Address3", "Address4"]
addresses2 = ["compAddress1", "compAddress2", "compAddress3", "compAddress4", "compAddress5"]

for a in addresses1:
    for b in addresses2:
        BLOCK OF CODE WITH GEOLOCATIONS

        if(distance == 1km):
            print("Addresses are within 1km from each other, move on next one")
            break

for a in addresses_1:
    for b in addresses_2:
        if distance(a, b) <= 1.00:
            print(a, b, "are within 1 km")
            break

---

您在引用的副本中遗漏了适用的解决方案。您需要

中断

内部循环；这将自然地延续外部循环：

addresses1 = ["Address1", "Address2", "Address3", "Address4"]
addresses2 = ["compAddress1", "compAddress2", "compAddress3", "compAddress4", "compAddress5"]

for a in addresses1:
    for b in addresses2:
        BLOCK OF CODE WITH GEOLOCATIONS

        if(distance == 1km):
            print("Addresses are within 1km from each other, move on next one")
            break

for a in addresses_1:
    for b in addresses_2:
        if distance(a, b) <= 1.00:
            print(a, b, "are within 1 km")
            break

对于地址_1中的地址：
对于地址2中的b：
如果距离（a，b）您从引用的副本中遗漏了适用的解决方案。您需要
中断
内部循环；这将自然地延续外部循环：

addresses1 = ["Address1", "Address2", "Address3", "Address4"]
addresses2 = ["compAddress1", "compAddress2", "compAddress3", "compAddress4", "compAddress5"]

for a in addresses1:
    for b in addresses2:
        BLOCK OF CODE WITH GEOLOCATIONS

        if(distance == 1km):
            print("Addresses are within 1km from each other, move on next one")
            break

for a in addresses_1:
    for b in addresses_2:
        if distance(a, b) <= 1.00:
            print(a, b, "are within 1 km")
            break

对于地址_1中的地址：
对于地址2中的b：
如果距离（a，b），你可能最好写两个循环，而不是尝试过滤产品。这是一样的，如果我像在其他编程语言中一样使用两个循环，则不起作用。你可能最好写两个循环，而不是尝试过滤产品。这是一样的，如果我像在其他编程语言中一样使用两个for循环，则不起作用。终于起作用了。真是难以置信。。。我无法想象一个多么愚蠢的小姐。我们大多数人都能想象——我们曾经去过那里。你觉得我是怎么认出这个的？：-）终于开始工作了。真是难以置信。。。我无法想象一个多么愚蠢的小姐。我们大多数人都能想象——我们曾经去过那里。你觉得我是怎么认出这个的？：-）