Python:对匹配元素

Question

我有以下数据结构：

a = [('customerA', '1.0.0'), ('customerB', '1.0.0'), ('customerC', '1.0.1')]
b = (('customerB', '1.1.0'), ('customerC', '1.0.1'))

我希望结果是这样的：

^{pr2}$

甚至完全跳过不存在的客户：

[('customerB', '1.0.0', '1.1.0'), ('customerC', '1.0.1', '1.0.1')]

在这种情况下，zip函数没有帮助，因为b来自客户名称的MySQLCursor.fetchall()withWHERE子句，因此如果客户不存在，它将与a不匹配：

>>> [a + (b[1],) for a, b in zip(a, b)]
[('customerA', '1.0.0', '1.1.0'), ('customerB', '1.0.0', '1.0.1')]
>>> import itertools
>>> for a, b in itertools.zip_longest(a, b):
...     print(a, b)
... 
('customerA', '1.0.0') ('customerB', '1.1.0')
('customerB', '1.0.0') ('customerC', '1.0.1')
('customerC', '1.0.1') None

Answer 1

使用collections。在

演示：

import collections
a = [('customerA', '1.0.0'), ('customerB', '1.0.0'), ('customerC', '1.0.1')]
b = (('customerB', '1.1.0'), ('customerC', '1.0.1'))

checkDict = dict(b)
d = collections.defaultdict(list)
for i in (a + list(b)):
    if i[0] in checkDict.keys():
        d[i[0]].append(i[1])
print(d)

输出：

^{pr2}$

Answer 2

In [11]: a = [('customerA', '1.0.0'), ('customerB', '1.0.0'), ('customerC', '1.0.1')]
    ...: b = (('customerB', '1.1.0'), ('customerC', '1.0.1'))

In [12]: ad = dict(a)

In [13]: bd = dict(b)

In [14]: [(k, ad.get(k), bd.get(k)) for k in set(ad.keys()) & set(bd.keys())]
Out[14]: [('customerC', '1.0.1', '1.0.1'), ('customerB', '1.0.0', '1.1.0')]

Answer 3

你试过直接做吗？在

customers_a = dict(a)
result = [(customer, customers_a[customer], version) for customer, version in b if customer in customers_a]

现在，result就是

^{pr2}$