使用python打印备用线

Question

I have a text file like

1
2
3
4
5
6
7
8
9
10

... and so on.

How do I write a program to print the first 2 lines then skip 3 then print 2 lines (that's the pattern.)

I'm a complete noob.
Any help will be appreciated.

thanks.

Answer 1

Based on line enumeration and assuming a 5-item cycle (display first two items, skip next three items):

for i, line in enumerate(file('myfile.txt')):

   if i % 5 in (0, 1):
       print line

Answer 2

I don't know if there's a pattern for this, really. You could just file.readlines() the entire thing and use array slicing. If you're concerned about memory consumption, iterate over the file handle using itertools.compress() on a pattern generated by itertools.cycle() . Or, you know, just write a loop or list comprehension.

Answer 3

The best that I can come up with is to use the modulo operator. Something like this:

f = open('filename.txt', 'r')
for index, line in enumerate(f.readlines()):
    if index%5 <= 1:
       print(line)

This should produce the pattern you are looking for.

Answer 4

Since others beat me to the answer here's the one line version:

print list(line for lineNum, line in enumerate(open("test.txt", "r")) if lineNum % 5 in (0, 1))

:-)

Answer 5

with open(filename) as f:
    print ''.join( f.readline() for i in xrange(7) if i in (0,1,5,6))

or

with open(filename) as f:
    print ''.join( f.readline() for i in '1100011' if i=='1')

or

with open(filename) as f:
    print ''.join( i*f.readline() for i in (1,1,0,0,0,1,1))

Answer 6

Building on @pynator's answer, and using the itertools module , here is the same solution as a ifilter - imap combination.

z is an iterable, like file("myfile.txt") . imap is used to pick the original data item from an enumerated pair.

>>> from itertools import ifilter, imap
>>> result = imap(lambda x: x[1], ifilter(lambda x: x[0]%5 in (0,1), enumerate(z)))
>>> for i in result: print i
... 
line 1
line 2
line 6
line 7
line 11
line 12
line 16
line 17
>>>