How to take only first four elements of every strings from a list

Question

I have a list in pyspark where list looks like

result = ['2016-12-11T04:12:58.797', '2016-12-11T03:50:28.253', '2016-12-11T03:49:52.613', '2016-12-11T03:37:49.857']

I have to fetch only the year from the list. What I tried is

resultYear = result[0:4]

But, I know this is not the solution. I am very new to python and pyspark, so I need help. Thanks.

Answer 1

To answer the question in the title, you simply have to iterate through the list and get the first 4 letters for each element of the list:

for element in result:
    year = element[:4] 
    # do what you want with this, e.g print it
    print(year)

>>>2016
>>>2016
...

But a more concise way to do it is list comprehension:

r = [el[:4] for element in result]
# returns a list of years

print(r)
>>> ['2016', '2016',...]

Answer 2

Use string split function and split the string where 'T' occurs and use the string before character 'T'
INPUT

result = ['2016-12-11T04:12:58.797', '2016-12-11T03:50:28.253', '2016-12-11T03:49:52.613', '2016-12-11T03:37:49.857']
result = [(r.split('-')[0]) for r in result]

OUTPUT

['2016', '2016', '2016', '2016']