Question

我有一个元组列表，我正在尝试根据最小值删除重复项：

a_list = [("1","111","15"),("2","111","10"),("3","111","5"),("4","112","40"),("5","112","10")]

Output : 
id id_client value2
1   111         15
2   111         10
3   111         5
4   112         40
5   112         10

Required Output
id id_client value2
3   111         5
5   112         10

我尝试了所有但都无法完成。

Answer 1

尝试以下代码：

# Input list
a_list = [("1","111","15"),("2","111","10"),("3","111","5"),("4","112","40"),("5","112","10")]

# Sort the list by the third value (index-2)
sorted_list = sorted(a_list, key=lambda x: int(x[2]))

# Track visited and repeated elements to only add the first tuple(x) with the smallest x[1]
visited = []

# New list to only append unique tuples
new_list = []

for i in sorted_list:
    if i[1] not in visited:
        new_list.append(i)
        visited.append(i[1])
print(new_list)

输出：

[('1', '111', '15'), ('4', '112', '40')]

Answer 2

尝试一下：

from itertools import groupby

new_l = []
for k,v in groupby(list, lambda x: x[1]):
    new_l.append(min(filter(lambda x:x[1]==k, list), key=lambda x:int(x[2])))

new_l将作为您的输出。

请注意，请勿将list之类的前缀名称用作变量名称。这些名字在python中意味着什么。

Answer 3

仅出于理解的乐趣而无需导入的另一种可能性：

lst = [("1","111","15"),("2","111","10"),("3","111","5"),("4","112","40"),("5","112","10")] 

[min((x for x in lst if x[1] == client), key=lambda x: int(x[2])) for client in {row[1] for row in lst}]

给予

 [('5', '112', '10'), ('3', '111', '5')]

在Python中搜索和过滤元组列表

3 个答案: