從列表python中刪除重復項

Question

我有兩個清單：

list1 = [IM12345, IM12346, IM12347, IM12348]
list2 = [ID300, ID404, ID300, ID601]

list2與相應的list1值相關聯。 list1具有唯一的值，而list2具有重復的值。

我要使list2唯一對應的關聯值將添加到該list2值中。

Dict= {ID300: {IM12345, IM12347}, ID404: IM12346, ID601: IM12348}

上面的圖案可以在列表，集合或字典中。

我應該使用python中的哪個算法來獲得以上結果？

Answer 1

您可以嘗試collections.defaultdict ：

from collections import defaultdict
d = defaultdict(set)

list1 = ['IM12345', 'IM12346', 'IM12347', 'IM12348']
list2 = ['ID300', 'ID404', 'ID300', 'ID601']

for key, value in zip(list2, list1):
    d[key].add(value)

演示：

>>> d
defaultdict(<class 'set'>, {'ID300': {'IM12345', 'IM12347'}, 'ID404': {'IM12346'}, 'ID601': {'IM12348'}})
>>>
>>>
>>> for i, j in d.items():
...     print(i, j)
...     
... 
ID601 {'IM12348'}
ID300 {'IM12345', 'IM12347'}
ID404 {'IM12346'}

Answer 2

您可以創建一個字典來保存數據集

list1 = ["IM12345", "IM12346", "IM12347", "IM12348"]
list2 = ["ID300", "ID404", "ID300", "ID601"]

dictResult=dict()
i=0
for item in list2:
    print item
    if dictResult.has_key(item):
        dictResult[item].append(list1[i])
    else:
        dictResult[item]=[list1[i]]
    i=i+1

print dictResult

結果：

{'ID404': ['IM12346'], 'ID300': ['IM12345', 'IM12347'], 'ID601': ['IM12348']}

Answer 3

可能不完全是pythonic的方式，但是-可以了：映射輸入：

map = dict(zip(list1, list2))

現在您可以進行逆映射：

inv_map = {}
for k, v in map.iteritems():
    inv_map[v] = inv_map.get(v, [])
    inv_map[v].append(k)

上面示例的結果：

>>> inv_map
{'ID404': ['IM12346'], 'ID300': ['IM12345', 'IM12347'], 'ID601': ['IM12348']}

Answer 4

做到這一點的另一種方法可能是使用列表操作。

yourList = ["1","2","3","4","1","2"] 
newList = [] 
for f in yourList:
    if f not in newList:
       newList.append(f)

Answer 5

簡單的解決方案

from collections import defaultdict

list1 = ['IM12345', 'IM12346', 'IM12347', 'IM12348']
list2 = ['ID300', 'ID404', 'ID300', 'ID601']

d = defaultdict(list)

for n in range(len(list2)):
    d[list2[n]].append(list1[n])

print d.items()

結果：

[('ID404', ['IM12346']), ('ID300', ['IM12345', 'IM12347']), ('ID601', ['IM12348'])]

Python2.7文檔---- defaultdict