Question

在这段代码中，我试图在字符串列表中找到最常用的名称，以使程序以O（nlogn）的身份运行。我知道这可以在O（n）中使用字典来完成。有什么明显的方法可以使这段代码更好？

def mostCommonName(L):
#in these two cases I'm making sure that if L has only one element in it
#or if it's empty that the correct result is returned before the loop
if len(L) == 1:
    return L[0]
if len(L) == 0:
    return None
#this automatically makes it nlogn
newL = sorted(L)
maxCount = 0
currCount = 1
maxName = set()
currName = newL[0]
for name in range(1,len(newL)):
    if newL[name] == currName:
        currCount += 1
    else:
        if currCount >= maxCount:
            maxCount = currCount
            currCount = 1
            maxName.add(currName)
            currName = newL[name]
        else:
            currCount = 1
            currName = newL[name]
if newL.count(newL[-1]) == maxCount:
    maxName.add(newL[-1])
if len(maxName) == 1:
    return maxName.pop()
return maxName

Answer 1

您可以改用groupby：

from operator import itemgetter
from itertools import groupby


def most_common_name(L):
    l = sorted(L)
    it = map(lambda pair: (pair[0], len(list(pair[1]))), groupby(l))
    r, _ = max(it, key=itemgetter(1))
    return r


result = most_common_name(['dan', 'david', 'dan', 'jen', 'james'])
print(result)

输出

dan

或更容易理解的替代方法：

from itertools import groupby


def len_group(pair):
    return sum(1 for _ in pair[1])


def most_common_name(l):
    sorted_l = sorted(l)
    r, _ = max(groupby(sorted_l), key=len_group)
    return r


result = most_common_name(['dan', 'david', 'dan', 'jen', 'james'])
print(result)

在这两种选择中，想法都是groupby处理连续值的分组。然后，您可以找到最大的组并返回该组的密钥。这些解决方案是 O（nlogn），如果您对 O（n）解决方案感兴趣，则可以使用Counter进行以下操作：

from operator import itemgetter
from collections import Counter


def most_common_name(l):
    counts = Counter(l)
    r, _ = max(counts.items(), key=itemgetter(1))
    return r


result = most_common_name(['dan', 'david', 'dan', 'jen', 'james'])
print(result)

输出

dan

Answer 2

稍微清洁一点，同时保持相同的算法：

def mostCommonName(L):
    if len(L) == 0:
        return None

    newL = sorted(L)
    occurrences, current_name = 1, newL[0]    
    best_occurrences, best_name = 1, newL[0]

    for i in range(1, len(newL)):
        if newL[i] == current_name:
            occurrences += 1
        else:
            if occurrences > best_occurrences:
                best_occurrences, best_name = occurrences, current_name
            occurrences = 1
            current_name = newL[i]
    return best_name

或者：

from collections import Counter

def mostCommonName(L):
    return Counter(L).most_common(1)[0][0]

提高算法的清晰度和效率

2 个答案: