Question

我有一个名单。它们按当年最受欢迎的女宝宝名字排序。在下面的函数中，我将报告在给定年份范围内连续出现的名称。我正在测试2004年到2007年的范围。

该列表如下：

['Emily','Emily','Emily','Isabel']

所需的输出是：

Emily occurred consecutively the most in this range at 3 time/s

但是我得到了

Emily occurred consecutively the most in this range at 2 time/s

我将在下面发布我的代码。

例子我被允许遵循。

['A', 'A', 'B', 'C', 'C', 'C']
bestName = 'A', bestCount = 1, currentName = 'A', currentCount = 1

['A', 'B', 'C', 'C', 'C'], next name = 'A'
currentName = nextName, so currentCount = 2

[ 'B', 'C', 'C', 'C'], next name = 'B'
currentName != nextName, and currentCount > bestCount, so bestName = 'A', bestCount = 2, 
now currentName = 'B', currentCount = 1

['C', 'C', 'C'] next name = 'C'
currentName != nextName, and currentCount not greater than bestCount
now currentName = 'C', currentCount = 1

['C', 'C'] next name = 'C'
currentName = nextName, so currentCount = 2

['C'] next name = 'C'
currentName = nextName, so currentCount = 3

[] no next name, loop ends
currentCount > bestCount, so bestName = 'C', bestCount = 3

代码：

def mostConsecutiveYears(names):
    """
    Compute which name occurs the most times consecutively in a
    list of names.
    :param names (list of Name): A list of name objects
    :return: A tuple containing best name (str) and the count (int)
    :rtype: tuple
    """
    bestName = names[0].name
    bestCount = 1
    currentName = names[0].name
    currentCount = 1

    for i in range(1,len(names)):
        if names[i].name == currentName:
            currentCount += 1

        if currentCount > bestCount:
            bestName = currentName
            bestCount = currentCount
            currentCount = 1
            currentName = names[i].name

    if currentCount > bestCount:
        bestName = currentName
        bestCount = currentCount

    return bestName, bestCount

Answer 1

看看itertools.groupby。这不是一个完整的解决方案，但应该让您了解如何继续。

names = ['a', 'a', 'a', 'b', 'c', 'c', 'a', 'b', 'c', 'c', 'c']
names = ['Emily', 'Emily', 'Emily', 'Isabel', 'Mary', 'Mary', 'Isabel', 'Emily']
groups = []
for a, b in itertools.groupby(names):
    items = list(b)
    groups.append((len(items), a))

count, name = sorted(groups, reverse=True)[0]
print("{} occurred consecutively the most in this range at {} time{}".format(name, count, "s" if count > 1 else ""))

输出

Emily occurred consecutively the most in this range at 3 times

真正的问题是你想如何处理像

这样的输入

['Emily', 'Isabel', 'Mary']

Answer 2

就像那样（注意names是我代码中的字符串列表）

def mostConsecutiveYears(names):
    """
    Compute which name occurs the most times consecutively in a
    list of names.
    :param names (list of Name): A list of name objects
    :return: A tuple containing best name (str) and the count (int)
    :rtype: tuple
    """
    bestName = names[0] # Be careful you can get an error if your list is empty
    bestCount = 1
    currentName = names[0]
    currentCount = 1

    for name in names[1:]:

        if name == currentName:
            currentCount += 1
        else:
            currentCount = 1
            currentName = name

        if currentCount > bestCount:
            bestName = currentName
            bestCount = currentCount

    return bestName, bestCount

print(mostConsecutiveYears(['Emily','Emily','Emily','Isabel']))

计算连续的名字

2 个答案: