基于元组的第二项从元组列表中删除列表项

简单明了的答案是:列表理解

ap = [    ('unknown', (81, 717, 236, 562)), 
          ('unknown', (558, 1033, 825, 765)), 
          ('unknown', (96, 1142, 225, 1013)), 
          ('Jenny', (558, 1033, 825, 765)), 
          ('unknown', (477, 1233, 632, 1078)), 
          ('unknown', (741, 1199, 868, 1070)), 
          ('Garry', (53, 282, 182, 153)), 
          ('Sam', (477, 1233, 632, 1078)), 
          ('Chen', (593, 283, 779, 97)), 
          ('Steve', (741, 1199, 868, 1070)), 
          ('unknown', (53, 282, 182, 153)), 
          ('Harry', (81, 717, 236, 562)), 
          ('unknown', (593, 283, 779, 97))]

known = [my_tuple[1] for my_tuple in ap if my_tuple[0] != "unknown"]
output = [my_tuple for my_tuple in ap if (my_tuple[1] in known and my_tuple[0] != "unknown") or my_tuple[1] not in known]

print(output)

[('unknown', (96, 1142, 225, 1013)), ('Jenny', (558, 1033, 825, 765)), ('Garry', (53, 282, 182, 153)), ('Sam', (477, 1233, 632, 1078)), ('Chen', (593, 283, 779, 97)), ('Steve', (741, 1199, 868, 1070)), ('Harry', (81, 717, 236, 562))]

这里发生的事情是,我们将所有的第二元组元素收集到一个列表中,其中的名称不是“未知”(使用列表理解)。

听起来可能有点混乱,希望你能理解我的意思。如果你有任何问题,请告诉我。

del 陈述

删除一个名称将从本地或本地数据库中删除该名称的绑定 全局命名空间,取决于名称是否出现在全局命名空间中 将引发异常。

用起来更好 dictionary 对于此任务。

expected = [('Harry',(81, 717, 236, 562)), ('Jenny', (558, 1033, 825, 765)),
('unknown', (96, 1142, 225, 1013)), ('Sam', (477, 1233, 632, 1078)),
('Steve', (741, 1199, 868, 1070)), ('Garry', (53, 282, 182, 153)), ('Chen', (593, 283, 779, 97))]


person_dict = {}

for person_name, person_val in ap:
    
    if person_val not in person_dict:
        # create key using tuple item 2
        person_dict[ person_val] =  person_name
    
    # key already exist so we only want to update its value if it is still unknown
    elif person_dict[ person_val] == 'unknown':
        person_dict[ person_val] =  person_name
        
        
ap = [(v,k) for k, v in person_dict.items()]

print(ap == expected) # True

在两个循环中使用同一个集合,并从循环中的集合中删除,这就产生了问题。我认为最好是创建一个新的项目列表,在这个循环之后删除它们。

toDelete = []
for i in ap:
    for j in ap:
        if i[1] == j[1] and not ap.index(i) == ap.index(j):
            if i[0] == "unknown":
                toDelete.append(i)
            else:
                toDelete.append(j)

for i in toDelete:
    try:
        ap.remove(i)
    except Exception as e:
        pass

可以通过将第二个循环看作:

for j in ap[ap.index(i)+1:]:

另一种方法是 collections 在原始列表中创建一个重复的列表,然后通过检查哪个元素是重复的来创建一个新的元组列表。然后创建一个新列表,检查哪个元素不重复,或者哪个元素不重复 unknown :

import collections

# Create a list of elements that are duplicate in the original list
duplicates = [item for item, count in collections.Counter([x[1] for x in ap]).items() if count > 1]


new = []
for elem in ap:
    if elem[1] in duplicates:
        if elem[0] != 'unknown':
            # Copy the duplicate element only if it's not unknown
            new.append(elem)
    else:

        new.append(elem)
print 'New list: ',new

输出为:

new list:  [('unknown', (96, 1142, 225, 1013)), ('Jenny', (558, 1033, 825, 765)), ('Garry', (53, 282, 182, 153)), ('Sam', (477, 1233, 632, 1078)), ('Chen', (593, 283, 779, 97)), ('Steve', (741, 1199, 868, 1070)), ('Harry', (81, 717, 236, 562))]