How to print the first n lines of file?

Question

I'm sure I'm missing something obvious and probably asked before but I can't seem to get the right combination of keywords together to give me an answer.

How can I write out the first n lines of a file (in effect, the opposite of file.readlines()[0:10])?

e.g. I have a function that takes in an input file, and needs to process information from the latter part, throwing out a header. However I want to keep the multi-line header, to be put back in to an output file.

def readInfile(infile):

    with open(infile, 'r') as ifh:
        # Skip exta info at top of file
        header = ifh.readline()[0:10] # Keep the header for later?

        noheader = ifh.readlines()[11:]
        for line in noheader:
        # Do the useful stuff
            usefulstuff = foo()

return usefulstuff, header

Then later I want to write out in the format of the input file, using their header:

print(header)
for thing in usefulstuff:
   print(thing)

Is there a method I'm missing, or is readlines no good for this as it returns a list?

I assumed

for line in header:
     print(line)

would work, but it doesn't seem to in this case - so I must be doing something wrong?

EDIT

Why does trying to use readlines()[] twice fail for the second range?

I fixed the code as @pbuck pointed out, that the header line should have been readlines() not readline but now the noheader variable is empty? Do I really have to open the file twice?!

Answer 1

Literally, read first n lines, then stop.

def read_first_lines(filename, limit):
  result = []
  with open(filename, 'r') as input_file:
    # files are iterable, you can have a for-loop over a file.
    for line_number, line in enumerate(input_file):
      if line_number > limit:  # line_number starts at 0.
        break
      result.append(line)
  return result