Python 为什么我不能删除列表中的某些值？ 匹配的_边=[0，-1,1，-1,2,3,3,2,4,6,5,7,6,4,7,5,8,9,9,8] 过程=匹配的_边 印刷工艺 对于匹配_边中的tup： 如果tup[1]，则进程中的tup[0]： process.removetup[1]，tup[0] 如果tup中为-1： 清除 打印图 印刷工艺

[0，-1,1，-1,2,3,3,2,4,6,5,7,6,4,7,5,8,9,9,8] 0, -1 [1，-1,2,3,4,6,5,7,8,9]

我尝试了很多方法，但我就是不能删除1，-1。怎么了

-1:1，-1 真的

---

你试过类似的东西吗

process = [tup for tup in matched_edges if (tup[0] != -1 and tup[1] != -1)]

这将删除任何包含-1的元组

编辑

另外，代码行为异常的原因是匹配的_边和进程都持有相同的对象，所以当您从进程中移除对象时，您也在迭代过程中从匹配的_边中移除它们，从而导致意外的行为。运行下面的代码，你就会明白我的意思

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = matched_edges
print(process)

for tup in matched_edges:
    print(tup)
    print("Length of matched edges:", len(matched_edges))
    if (tup[1], tup[0]) in process:
        process.remove((tup[1], tup[0]))
    if -1 in tup:
        process.remove(tup)
        print("removing", tup)
print(process)

matched_edges = [tup for tup in matched_edges if (tup[0] != -1 and tup[1] != -1)]
print(matched_edges)

---

你试过类似的东西吗

process = [tup for tup in matched_edges if (tup[0] != -1 and tup[1] != -1)]

这将删除任何包含-1的元组

编辑

另外，代码行为异常的原因是匹配的_边和进程都持有相同的对象，所以当您从进程中移除对象时，您也在迭代过程中从匹配的_边中移除它们，从而导致意外的行为。运行下面的代码，你就会明白我的意思

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = matched_edges
print(process)

for tup in matched_edges:
    print(tup)
    print("Length of matched edges:", len(matched_edges))
    if (tup[1], tup[0]) in process:
        process.remove((tup[1], tup[0]))
    if -1 in tup:
        process.remove(tup)
        print("removing", tup)
print(process)

matched_edges = [tup for tup in matched_edges if (tup[0] != -1 and tup[1] != -1)]
print(matched_edges)

---

您可以创建一个新列表

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = []
for tup in matched_edges:
    if (tup[1], tup[0]) in matched_edges:matched_edges.remove((tup[1], tup[0]))
    if -1 not in tup:process.append(tup)
print(process)

输出

---

您可以创建一个新列表

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = []
for tup in matched_edges:
    if (tup[1], tup[0]) in matched_edges:matched_edges.remove((tup[1], tup[0]))
    if -1 not in tup:process.append(tup)
print(process)

输出

---

这是错误的线路

process = matched_edges

这是一条编程规则。尽可能不要更新当前正在迭代的对象！代码中所做的是，进程和匹配的_边指向/引用堆内存中相同的对象/列表不同的堆栈变量，但都指向堆中相同的内存地址，即实际的列表。因此，更新进程意味着您也在间接更新匹配的_边

您当前正在迭代匹配的_边：

for tup in matched_edges:

process.remove(...)

然后，您更新了流程，这会打乱匹配_边的迭代：

for tup in matched_edges:

process.remove(...)

对代码的更正是使用以下内容：

process = matched_edges.copy()

---

这是错误的线路

process = matched_edges

这是一条编程规则。尽可能不要更新当前正在迭代的对象！代码中所做的是，进程和匹配的_边指向/引用堆内存中相同的对象/列表不同的堆栈变量，但都指向堆中相同的内存地址，即实际的列表。因此，更新进程意味着您也在间接更新匹配的_边

您当前正在迭代匹配的_边：

for tup in matched_edges:

process.remove(...)

然后，您更新了流程，这会打乱匹配_边的迭代：

for tup in matched_edges:

process.remove(...)

对代码的更正是使用以下内容：

process = matched_edges.copy()

此行过程=匹配的_边实际上是创建对匹配的_边列表的引用，而不是项目的副本。反过来，这会影响for循环中的迭代器，因为您要从进程和匹配的_边中删除项。为了演示，我稍微修改了您的代码，只是为了打印值：

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = matched_edges

for tup in matched_edges:
    print(f"Current tup: {tup}")
    print(f"Process: {process}")
    print(f"Matched: {matched_edges}")
    if (tup[1], tup[0]) in process:
        process.remove((tup[1], tup[0]))
    if -1 in tup:
        process.remove(tup)

输出：

Current tup: (0, -1)
Process: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (2, 3)
Process: [(1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (4, 6)
Process: [(1, -1), (2, 3), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (5, 7)
Process: [(1, -1), (2, 3), (4, 6), (5, 7), (7, 5), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (4, 6), (5, 7), (7, 5), (8, 9), (9, 8)]
Current tup: (8, 9)
Process: [(1, -1), (2, 3), (4, 6), (5, 7), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (4, 6), (5, 7), (8, 9), (9, 8)]

您可以看到，在第二次迭代中，您最终跳过了1，-1元组。通常，每次删除项目时都会跳过该项目

将作业更改为“处理”以使用“复制”，以便获得新列表：

process = matched_edges.copy()

这应该可以纠正这种奇怪的行为。虽然我不太确定您使用的循环是否仍然完全符合您的预期，但这是复制列表的输出：

Current tup: (0, -1)
Process: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (1, -1)
Process: [(1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (2, 3)
Process: [(2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (3, 2)
Process: [(2, 3), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (4, 6)
Process: [(4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (5, 7)
Process: [(4, 6), (5, 7), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (6, 4)
Process: [(4, 6), (5, 7), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (7, 5)
Process: [(5, 7), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (8, 9)
Process: [(8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (9, 8)
Process: [(8, 9)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]

此行过程=匹配的_边实际上是创建对匹配的_边列表的引用，而不是项目的副本。反过来，这会影响for循环中的迭代器，因为您要从进程和匹配的_边中删除项。为了演示，我稍微修改了您的代码，只是为了打印值：

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = matched_edges

for tup in matched_edges:
    print(f"Current tup: {tup}")
    print(f"Process: {process}")
    print(f"Matched: {matched_edges}")
    if (tup[1], tup[0]) in process:
        process.remove((tup[1], tup[0]))
    if -1 in tup:
        process.remove(tup)

输出：

Current tup: (0, -1)
Process: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (2, 3)
Process: [(1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (4, 6)
Process: [(1, -1), (2, 3), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (5, 7)
Process: [(1, -1), (2, 3), (4, 6), (5, 7), (7, 5), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (4, 6), (5, 7), (7, 5), (8, 9), (9, 8)]
Current tup: (8, 9)
Process: [(1, -1), (2, 3), (4, 6), (5, 7), (8, 9), (9, 8)]
Matched: [(1, -1), (2, 3), (4, 6), (5, 7), (8, 9), (9, 8)]

您可以看到，在第二次迭代中，您最终跳过了1，-1元组。通常，每次删除项目时都会跳过该项目

将作业更改为“处理”以使用“复制”，以便获得新列表：

process = matched_edges.copy()

这应该可以纠正这种奇怪的行为。虽然我不太确定您使用的循环是否仍然完全符合您的预期，但这是复制列表的输出：

Current tup: (0, -1)
Process: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (1, -1)
Process: [(1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (2, 3)
Process: [(2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (3, 2)
Process: [(2, 3), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (4, 6)
Process: [(4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (5, 7)
Process: [(4, 6), (5, 7), (7, 5), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (6, 4)
Process: [(4, 6), (5, 7), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (7, 5)
Process: [(5, 7), (8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (8, 9)
Process: [(8, 9), (9, 8)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
Current tup: (9, 8)
Process: [(8, 9)]
Matched: [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]

---

问题是，如果您执行process=matched_edges，那么您正在创建一个变量流程，该流程指向与matched_edges相同的对象，以创建一个具有相同值的新列表，您应该改为执行process=matched_edges[：]

因此，您的代码如下所示：

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = matched_edges[:]
print(process)

for tup in matched_edges:
    if (tup[1], tup[0]) in process:
        process.remove((tup[1], tup[0]))
    if -1 in tup:
        process.remove(tup)
        print(tup)
print(process)

---

问题是，如果您执行process=matched_edges，那么您正在创建一个变量流程，该流程指向与matched_edges相同的对象，以创建一个具有相同值的新列表，您应该改为执行process=matched_edges[：]

因此，您的代码如下所示：

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]
process = matched_edges[:]
print(process)

for tup in matched_edges:
    if (tup[1], tup[0]) in process:
        process.remove((tup[1], tup[0]))
    if -1 in tup:
        process.remove(tup)
        print(tup)
print(process)

---

以下内容对我来说很好

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), 
                (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]

process = []

for tup in matched_edges:
    if (tup[0] != -1 and tup[1] != -1) and (tup[1], tup[0]) not in process:
        process.append(tup)
print(process)

输出

[(2, 3), (4, 6), (5, 7), (8, 9)]

---

以下内容对我来说很好

matched_edges = [(0, -1), (1, -1), (2, 3), (3, 2), (4, 6), 
                (5, 7), (6, 4), (7, 5), (8, 9), (9, 8)]

process = []

for tup in matched_edges:
    if (tup[0] != -1 and tup[1] != -1) and (tup[1], tup[0]) not in process:
        process.append(tup)
print(process)

输出

[(2, 3), (4, 6), (5, 7), (8, 9)]

---

移除元组必须满足什么条件？当元组中的-1或顺序相同但不同的元组时。如1,2和2,1，则我需要移除其中一个。这是否回答了您的问题？

---

请记住，过程和匹配的_边是对同一列表的引用，因此，当从列表中删除时，您在列表上方。根据您的解释@NiceChang当元组中的-1或反向元组存在时，您的整个列表似乎将被删除？删除元组必须满足什么条件？当元组中的-1或相同但顺序不同的元组时。如1,2和2,1，然后我需要移除其中一个。这能回答你的问题吗？请记住，进程和匹配的_边是对同一列表的引用，因此在从列表中删除时，您处于列表上方。根据您的解释@NiceChang，当-1在元组或反向元组中存在时，您的整个列表似乎都将被删除？请告诉我：请告诉我：