你如何让python在列表中找到特定的文本然后打印出来

Question

This is what I have so far:

def lists():
    global ClientList, highList, moderateList
    ClientList = [    ["NeQua,High"],
                      ["ImKol,Moderate"],
                      ["YoTri,Moderate"],
                      ["RoDen,High"],
                      ["NaThe,Moderate"],
                      ["ReWes,Moderate"],
                      ["BrFre,High"],
                      ["KaDat,High"],
                      ["ViRil,High"],
                      ["TrGeo,High"]]
     highList = ["Running", "Swimming", "Aerobics", "Football", "Tennis"]
     moderateList = ["Walking", "Hicking", "Cleaning", "Skateboarding", "Basketball"]
     checkclient()

def checkclient():
    global ClientList, highList, moderateList
    answer = input("please enter the client ID: ")
    answer2 = next(answer for answer in ClientList)
    print(answer)

So I want to input the specific clientID, I want python to then find the client ID in the list, print the clientID with the intensity level (high or moderate) so I can use it later to ask the user how many minutes they spent exercising in the different activities based on whether their intensity was high or moderate.

At the moment the code only prints the first part of the list regardless of what the variable answer is: ["NeQua, High"].

Please can you tell me how to fix this and try to keep it simple as I am relatively new to Python.

Thanks Cameron

Answer 1

Use a dictionary instead (and there's no need to wrap it in a function that does nothing but create global objects).

ClientList = {"NeQua":"High",
              "ImKol":"Moderate",
              "YoTri":"Moderate",
              "RoDen":"High",
              "NaThe":"Moderate",
              "ReWes":"Moderate",
              "BrFre":"High",
              "KaDat":"High",
              "ViRil":"High",
              "TrGeo":"High"}

You don't need to specify mutable objects like list s or dictionaries as global if all you want to do is mutate them. You only need global if you want local assignments to the same name to also assign to the global name. More importantly, though, next() just returns the next element in an iterable. As a list is an ordered sequence, a generator that you make out of it with answer for answer in ClientList will have the same order, and the next() of that (redundant, I might add) generator will always be the first element of ClientList , because you keep making a new generator. If you want next() to proceed through the whole thing, you'd have to save it first. However, none of that is necessary here. Just access the dictionary. I use get() here to avoid errors if the user tries to access a user that doesn't exist.

def checkclient():
    answer = input("please enter the client ID: ")
    print(ClientList.get(answer, 'Not found.'))

checkclient()

Also note that a function must be defined before it is called (order matters).

Answer 2

You might change it as follows:

def checkclient():
    global ClientList, highList, moderateList
    answer = input("please enter the client ID: ")  # input: NaThe
    try:
        answer2 = next(a for a in ClientList if answer in a) 
    except StopIteration:
        checkclient()
    else:
        print(answer2)  # NaThe,Moderate

next returns the first element of an iterable so you always get the first element of ClientList , so you need to filter out the ones containing your ID.