How would I go about getting the first N lines of a text file in python? With N have to give as argument
usage:
python file.py datafile -N 10
My code
import sys
from itertools import islice
args = sys.argv
print (args)
if args[1] == '-h':
print ("-N for printing the number of lines: python file.py datafile -N 10")
if args[-2] == '-N':
datafile = args[1]
number = int(args[-1])
with open(datafile) as myfile:
head = list(islice(myfile, number))
head = [item.strip() for item in head]
print (head)
print ('\n'.join(head))
I wrote the program, can let me know better than this code
Assuming that the print_head logic you've implemented need not be altered, here's the script I think you're looking for:
import sys
from itertools import islice
def print_head(file, n):
if not file or not n:
return
with open(file) as myfile:
head = [item.strip() for item in islice(myfile, n)]
print(head)
def parse_args():
result = {'script': sys.argv[0]}
args = iter(sys.argv)
for arg in args:
if arg == '-F':
result['filename'] = next(args)
if arg == '-N':
result['num_lines'] = int(next(args))
return result
if __name__ == '__main__':
script_args = parse_args()
print_head(script_args.get('filename', ''), script_args.get('num_lines', 0))
Running the script
python file.py -F datafile -N 10
Note: The best way to implement it would be to use argparse
library
Arguments are available via the sys
package.
Example 1 : ./file.py datafile 10
#!/usr/bin/env python3
import sys
myfile = sys.argv[1]
N = int(sys.argv[2])
with open("datafile") as myfile:
head = myfile.readlines()[0:args.N]
print(head)
Example 2 : ./file.py datafile --N 10
If you want to pass multiple optional arguments you should have a look at the argparse package.
#!/usr/bin/env python3
import argparse
parser = argparse.ArgumentParser(description='Read head of file.')
parser.add_argument('file', help='Textfile to read')
parser.add_argument('--N', type=int, default=10, help='Number of lines to read')
args = parser.parse_args()
with open(args.file) as myfile:
head = myfile.readlines()[0:args.N]
print(head)
You can access argument passed to the script through sys
sys.argv
The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string.
So in code it would look like this:
import sys
print("All of argv")
print(sys.argv)
print("Last element every time")
print(sys.argv[-1])
Reading the documentation you'll see that the first values stored in the sys.argv vary according to how the user calls the script. If you print the code I pasted with different types of calls you can see for yourself the kind of values stored.
For a basic first approach: access n through sys.argv[-1] which returns the last element every time, assuming. You still have to do a try and beg for forgiveness to make sure the argument passed is a number. For that you would have:
import sys
try:
n = int(sys.argv[-1])
except ValueError as v_e:
print(f"Please pass a valid number as argument, not ${sys.argv[-1]}")
That's pretty much it. Obviously, it's quite basic, you can improve this even more by having the users pass values with flags, like --skip-lines 10 and that would be your n, and it could be in any place when executing the script. I'd create a function in charge of translating sys.argv into a key,value dictionary for easy access within the script.
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