如何迭代到列表中的元组

Question

I am writing a program where I have a bunch of tuples in a list like this:

[('a01', '01-24-2011', 's1'), ('a03', '01-24-2011', 's2') etc.

The tuples are the format

(animal ID, date(month, day, year), station# )

I do not know how to access the information about the month only.

I have tried:

months = []    
for item in list:
    for month in item:
        if month[0] not in months:
            months.append(month[0])

I am working in python 3.

Answer 1

L = [('a01', '01-24-2011', 's1'), ('a03', '01-24-2011', 's2')]
for animal, date, station in L:
    month, day, year = date.split('-')
    print("animal ID {} in month {} at station {}".format(animal, month, station))

Output:

animal ID a01 in month 01 at station s1
animal ID a03 in month 01 at station s2

Answer 2

The basic idea is to get the second item of the tuple, which is a string, then get the first two characters of the string. Those characters describe the month.

I'll go through the process step by step.

Let's say you have a list called data :

data = [('a01', '01-24-2011', 's1'), ('a03', '01-24-2011', 's2')]

Take the first item:

item = data[0]

The value of item is the tuple ('a01', '01-24-2011', 's1') .

Take the second element of item :

date = item[1]

The value of date is the string '01-24-2011' .

Take the first two characters of date :

month = date[:2]

The value of month is the string 01 . You can convert this into an integer:

month = int(month)

Now the value of month is 1 .

Answer 3

Using list comprehensions:

data = [('a01', '01-24-2011', 's1'), ('a03', '01-24-2011', 's2')]

months = [item[1].split('-')[0] for item in data]

print(months)

Answer 4

>>> my_list = [('a01', '01-24-2011', 's1'), ('a03', '01-24-2011', 's2')]
>>> [ x for x in map(lambda x:x[1].split('-')[0],my_list) ]
['01', '01']

you can use map and lambda

Answer 5

如果您只想要一个独特的月份列表并且订单无关紧要使用一套：

months = list({date.split("-",1)[0] for _, date, _ in l})