Python 在每个键的值处删除每个元素

我们如何比较i.year和j.year中的每个元素，如果元素彼此相等，则删除值中该索引处的元素 在每个关键点和，然后移动到下一个元素并继续该过程。每个值的长度将始终相同。不使用进口商品

所以基本上我想说的是：

为了解决这个问题，只比较i.year和j.year中的元素：

sample_dict = {'i.year': ['1997', '1997'], 'i.month': ['March', 'April'], 'j.month': ['March', 'April'], 'j.year': ['1997', '2003']}

我们得到：

-> {'i.year': ['1997', '1997'], 'i.month': ['March', 'April'], 'j.month': ['March', 'April'], 'j.year': ['1997', '2003']}
-> The first element in i.year is equal to the first element in j.year, so delete the first element in each value in every key.

-> {'i.year': ['1997', '1997', '2009'], 'i.month': ['March', 'April', 'June'], 'j.month': ['March', 'April', 'June'], 'j.year': ['1997', '2003', '2010']}
-> The first element in i.year is equal to the first element in j.year, so delete the first element in each value in every key.

-> {'i.year': ['1996', '1997', '2010'], 'i.month': ['March', 'April', 'June'], 'j.month': ['March', 'April', 'June'], 'j.year': ['1997', '2003', '2010']}
-> The first element in i.year is not equal to the first element in j.year, so we move on to the next element.
-> '1997' does not equal to '2003' so we move on to the next element
-> '2010' is equal to '2010' so we delete each element in every key at the index

另一个例子：

-> {'i.year': ['1997'], 'i.month': ['April'], 'j.month': ['April'], 'j.year': ['2003']}
-> The element now in i.year and j.year is the same so we're done. Return that dictionary

我们得到：

-> {'i.year': ['1997', '1997'], 'i.month': ['March', 'April'], 'j.month': ['March', 'April'], 'j.year': ['1997', '2003']}
-> The first element in i.year is equal to the first element in j.year, so delete the first element in each value in every key.

-> {'i.year': ['1997', '1997', '2009'], 'i.month': ['March', 'April', 'June'], 'j.month': ['March', 'April', 'June'], 'j.year': ['1997', '2003', '2010']}
-> The first element in i.year is equal to the first element in j.year, so delete the first element in each value in every key.

-> {'i.year': ['1996', '1997', '2010'], 'i.month': ['March', 'April', 'June'], 'j.month': ['March', 'April', 'June'], 'j.year': ['1997', '2003', '2010']}
-> The first element in i.year is not equal to the first element in j.year, so we move on to the next element.
-> '1997' does not equal to '2003' so we move on to the next element
-> '2010' is equal to '2010' so we delete each element in every key at the index

最后一个例子：

-> {'i.year': ['1997', '2009'], 'i.month': ['April', 'June'], 'j.month': ['April', 'June'], 'j.year': ['2003', '2010']}
-> Now the first element in i.year and j.year are different so we move to the next element which is '2009' and '2010'
-> '2009' and '2010' are different so we move to the next element, since there's none we are done. Return that dictionary.

我们得到：

-> {'i.year': ['1997', '1997'], 'i.month': ['March', 'April'], 'j.month': ['March', 'April'], 'j.year': ['1997', '2003']}
-> The first element in i.year is equal to the first element in j.year, so delete the first element in each value in every key.

-> {'i.year': ['1997', '1997', '2009'], 'i.month': ['March', 'April', 'June'], 'j.month': ['March', 'April', 'June'], 'j.year': ['1997', '2003', '2010']}
-> The first element in i.year is equal to the first element in j.year, so delete the first element in each value in every key.

-> {'i.year': ['1996', '1997', '2010'], 'i.month': ['March', 'April', 'June'], 'j.month': ['March', 'April', 'June'], 'j.year': ['1997', '2003', '2010']}
-> The first element in i.year is not equal to the first element in j.year, so we move on to the next element.
-> '1997' does not equal to '2003' so we move on to the next element
-> '2010' is equal to '2010' so we delete each element in every key at the index

我有这个想法，但无法将其转化为python代码：

对于i.month中的每个元素，通过j.year中的每个元素循环，如果元素不相等，则通过整个字典循环并删除该元素

---

在该索引处。

您可以首先找到

i.year

和

j.year

中元素不同的索引，然后迭代dict并过滤掉不在这些索引处的项目：

-> {'i.year': ['1996', '1997'], 'i.month': ['March', 'April'], 'j.month': ['March', 'April'], 'j.year': ['1997', '2003']}
-> There are no more elements to move on to, so we are done, we return this dictionary.

演示：

def solve(d, *keys):
    indexes = [i for i, x in enumerate(zip(*(d[k] for k in keys)))
                                                                if len(set(x)) != 1]
    return {k:[v[x] for x in indexes] for k, v in d.items()}

---

谢谢，Ashwini，但是有没有办法不用zip就可以做到这一点？@Sc4r是的，你可以，然后你必须写你的，这在这里是不必要的。哦，谢谢，我应该在你的代码中修改哪一行而不是我们拥有的那一行更大。