从列表中创建元组或新列表

Question

I have a list which comes from a text file that I have parsed using very primitive regular expressions. I would like to reorganize a more spartan list that contains only files with a date immediately following. I've tried looping through the list using len() but that will only extract the files and not the next entry. Many thanks in advance.

This:

2014-01-28

part002.csv.gz

2014-01-28

part001.csv.gz

2014-01-28

2014-01-28

2014-01-27

2014-01-27

2014-01-26

2014-01-26

2014-01-25

part002.csv.gz

2014-01-25

Becomes this:

part002.csv.gz

2014-01-28

part001.csv.gz

2014-01-28

part002.csv.gz

2014-01-25

Answer 1

You can use a list comprehension:

filtered = [e for i, e in enumerate(l) if not isDate(e) or (i > 0 and not isDate(l[i-1]))]

Complete example:

l = ['2014-01-28', 'part002.csv.gz', '2014-01-28', 'part001.csv.gz', '2014-01-28', '2014-01-28', '2014-01-27', 'part002.csv.gz', '2014-01-25']

def isDate (s):
    return '.' not in s

filtered = [e for i, e in enumerate(l) if not isDate(e) or (i > 0 and not isDate(l[i-1]))]

print (filtered)

---

Explained:

l is our original list.

isDate takes a string and tests whether it is a date (in my simple example it just checks that it doesn't contain a period, for better results use regex or strptime).

enumerate enumerates a list (or anything iterable, I will now stick to the word list , just in order not to get too technical). It returns a list of tuples; each tuple containing the index and the element of the list passed to enumerate. For instance enumerate (['a', None, 3]) makes [(0,'a'),(1,None),(2,3)]

i, e = unpacks the tuple, assigning the index to i and the element to e .

A list comprehension works like this (simplyfied): [x for x in somewhere if cond(x)] returns a list of all elements of somewhere which comply with the condition cond(x) .

In our case we only add elements to our filtered list, if they are no dates (not the fruit) not isDate(e) or if they are not at the beginning i > 0 and at the same time their predecessor is not a date not isDate(l[i-1]) (that is, a file).

In pseudocode:

Take list `l`
Let our filtered list be an empty list
For each item in `l` do
    let `i` be the index of the item
    let `e` be the item itself

    if `e` is not a Date
      or if `i` > 0 (i.e. it is not the first item)
      and at the sametime the preceding item is a File
      then and only then add `e` to our filtered list.

Answer 2

Store the previous line at each line, then you always have have it when you need it

previous_line = None
newlist = []
for line in lines:
    if isdate(line):
        newlist.append(previous_line)
    previous_line = line

Defining isdate :

import datetime
def isdate(s):
    try:
        datetime.datetime.strptime(s, '%Y-%m-%d')
    except:
        return False
    else:
        return True

Answer 3

Working through it:

s = """
#that long string, snipped
"""

li = [x for x in s.splitlines() if x]

li
Out[3]: 
['2014-01-28',
 'part002.csv.gz',
 '2014-01-28',
 'part001.csv.gz',
 '2014-01-28',
 '2014-01-28',
 '2014-01-27',
 '2014-01-27',
 '2014-01-26',
 '2014-01-26',
 '2014-01-25',
 'part002.csv.gz',
 '2014-01-25']

[tup for tup in zip(li,li[1:]) if 'csv' in tup[0]] #shown for dicactic purposes, gen expression used below
Out[7]: 
[('part002.csv.gz', '2014-01-28'),
 ('part001.csv.gz', '2014-01-28'),
 ('part002.csv.gz', '2014-01-25')]

The actual answer:

from itertools import chain

list(chain.from_iterable(tup for tup in zip(li,li[1:]) if 'csv' in tup[0]))
Out[9]: 
['part002.csv.gz',
 '2014-01-28',
 'part001.csv.gz',
 '2014-01-28',
 'part002.csv.gz',
 '2014-01-25']

Essentially: zip (in python 2, use izip ) the list together with itself, one index advanced. Iterate over the pairwise tuples, filtering out those that don't have a file-like string for their first element. Lastly, flatten the tuples into a list using itertools.chain to achieve your desired output.