简化您的逻辑：

def firstTeamPoints():
    firstTeamTotal.append(5 - int(firstTeamPlace))

如果事实证明它更复杂，你可以使用一个Map：

m = {
    "1": 4,
    "2": 3,
    "3": 2,
    "4": 1,
}
def firstTeamPoints():
    firstTeamTotal.append(m[firstTeamPlace])

您可以以更快、更灵活的方式完成您想做的事情：

#Some preparation. You can do it manually or with another function

tnames = ['team 1', 'team 2', 'team 3', 'team 4', 'team 5'] 
teamsPlaces = [] # This is for where the teams placed for each event, 1st, 2nd, 3rd etc
teamsTotals = [] # This is where the scores are added to make a final score for each team

#Here we prepare the final scores, with everyone starting from 0
for team in range(len(tnames)):
    teamsTotals.append(0)

def tevent1():
    print("This is the relay race event")
    print("Please enter the where each team placed next to their name (1-4)")
    eventTeamsPlaces = []
    for i in range(len(tnames)):
        teamPos = input(tnames[i])
        eventTeamsPlaces.append(teamPos) #for each event, you will have a list of places
        teamsTotals[i] = teamsTotals[i] + (len(tnames) - int(teamPos))
        
    teamsPlaces.append(eventTeamsPlaces) #and the list of places is appended to the list of places for all the events
    print(teamsTotals)
tevent1()

这样，在两个事件结束后，您的输出为：

teamsPlaces = [['1', '2', '3', '4', '5'], ['5', '4', '3', '2', '1']]
teamsTotals = [4, 4, 4, 4, 4]

end如果您想获得事件编号2的地点列表，您可以使用teamsPlaces[1]，即['5', '4', '3', '2', '1']
通过这种方法，您可以对所有事件使用这个简单的函数，而对团队数量和事件数量没有任何限制。