Question

我正在尝试比较两组匹配数字。一旦找到一个数字，就会将与数字相关联的用户名附加到字典以及它们匹配的数字量。我的解释不是最好的，但也许我的代码会显示我想要做的事情。

c.execute('SELECT * FROM {} WHERE drawnumber = (%s)'.format(numbertable),
          drawnumber)
numbers = c.fetchall()[0]
print(numbers)

numbers = set(imap(str.rstrip, numbers))

print(numbers)

c.execute('SELECT * FROM {} WHERE paid = (%s)'.format(table), paid)
entries = c.fetchall()
print(entries)

results = defaultdict(list)

for row in entries:
    user = row[1]
    number1 = row[2]
    number2 = row[3]
    number3 = row[4]
    number4 = row[5]
    number5 = row[6]
    pub_key = row[7]

    results[sum(n in numbers for n in row)].append(user)

仅供参考：设定数字如下：set(['11', '12', '15', '1', '3', '8'])

条目集看起来像这样：

((9L, 'test8', 12L, 4L, 3L, 15L, 8L, '1Bu1hXvk7NzoCgRFyXcGwFULtdSRsDNoCy', 'T'),
 (10L, 'test9', 12L, 4L, 1L, 15L, 8L, '1Bu1hXvk7NzoCgRFyXcGwFULtdSRsDNoCy', 'T'))

Answer 1

您可以使用set.intersection

for _, user, number1, number2, number3, number4, \
        number5, pub_key, _ in row:
    row_nums = set([number1, number2, number3, number4, number5])
    matching = row_nums.intersection(numbers)
    if matching:
        results[len(matching)].append(user)

但是numbers是一组字符串，而不是一组整数。你应该解决这个问题。

numbers = set(imap(int, imap(str.rstrip, numbers)))

比较两组数字

1 个答案: