如何获得最少使用的物品？

Question

假设你有一个使用计数的默认dict，如下所示：

usage_counts = collections.defaultdict(int)
usage_counts['foo1'] = 3
usage_counts['foo2'] = 3
usage_counts['foo3'] = 1
usage_counts['foo4'] = 1
usage_counts['foo5'] = 56
usage_counts['foo6'] = 65

在一些列表中有候选者foo1、foo3、foo4和foo5：

candidates = ['foo1', 'foo3', 'foo4', 'foo5']

如何从最少使用的候选人中随机挑选

我想出了这个函数，但我想知道是否有更好的方法

def get_least_used(candidates, usage_counts):
    candidate_counts = collections.defaultdict(int)
    for candidate in candidates:
        candidate_counts[candidate] = usage_counts[candidate]
    lowest = min(v for v in candidate_counts.values())
    return random.choice([c for c in candidates if candidate_counts[c] == lowest])

Answer 1

如果您接受显式地浏览列表，则只能浏览一次，方法是为具有最低计数的canditate生成一个列表。如果当前计数小于旧最小值，则初始化新列表，如果等于，则添加到列表：

def get_least_used(candidates, usage_counts):
    mincount = sys.maxint
    for c in candidates :
        count = usage_counts[c]
        if count < mincount:
            leastc = [ c ]
            mincount = count
        elif count == mincount:
            leastc.append(c)
    return random.choice(leastc)

正如您所说的使用python2.6，我用sys.maxint初始化mincount。在python3.x下，您必须选择一个合理的值great

Answer 2

random.shuffle(candidates)

min_candidate = min(candidates, key=usage_counts.get)

返回洗牌候选列表中的第一个“最小”候选