我有一个完整的程序,它有一个不同功能的前一个字典,它给我一个飞机的出发和到达城市列表。
我正在尝试编写一个函数来确定哪些键具有最多的传出航班,而我无法弄清楚如何找到哪些键具有最多的值。我的字典被命名为航班,其中出发城市为钥匙,抵达时为值。
def out(flight):长度= 0表示i in(flight):if(len(flights [i])> length):length =(len(flights [i]))break else:continue
for i in flights:
if (len(flights[i]) == length):
pop = (len(flights[i]))
print ("the most outgoing flight is: " , [i])
这段代码可能会起作用,但由于某种原因,它没有从文件中给出正确的最大输出。任何想法为什么?
最简单的解决方案是使用内置的max函数和列表理解:
def outgoing(flights):
print(max([len(i) for i in flights]))
如果您想坚持使用代码,则需要与每次迭代的最大值进行比较:
def outgoing(flights):
max_outgoing = 0
for i in flights:
if(max_outgoing < len(flights[i])):
print(max_outgoing)
max_outgoing = len(flights[i])
编辑:重新阅读你的问题,似乎你也想得到最大值的关键。这样做:
def outgoing(flights):
max_outgoing = 0
max_key = None
for i in flights:
if(max_outgoing < len(flights[i])):
print(max_outgoing)
max_outgoing = len(flights[i])
max_key = i
或者在较短的版本中:
def outgoing(flights):
out_dict = {i: len(i) for i in flights}
max_out = max(out_dict, key=out_dict.get)
print(max_out)
print(flights[max_out])
你不是很清楚flights
的结构,所以我假设它是键是字符串,值是字符串列表。
一种方法是创建一个元组列表,其中每个元素都是(departure_city, len(flights[departure_city]))
。然后,您可以按到达次数对列表进行排序。
def outgoing(flights):
# Create a list of tuples
flight_tups = [(departure_city, len(flights[departure_city])) for departure_city in flights]
# Sort the list by number of arrivals
# We do this by passing a lambda to `sort`,
# telling it to sort by the second value in
# each tuple, i.e. arrivals
flight_tups.sort(key=lambda tup: tup[1])
# We can now get the city with the most arrivals by
# taking the first element of flight_tups
most = flight_tups[0]
print(f'City: {most[0]}\nNumber of Arrivals: {most[1]}')
注意:你也可以使用max
,但是根据你的问题,你似乎想要拥有最多人数的城市,而不仅仅是最多的城市。使用sort
也可以让你知道是否有平局。