Read only first column from txt file

Question

I have simple txt file with maybe 6 columns and 10000 rows. How can i read only first column and save it in some array?

Every solution that i tried didn't work.

I manage it with octave very simply,

data = load("sample.txt") 
first_column = data(:,1)

EDIT: I forgot to metion, I have to rows of strings that i need to skip. My file starts like this:

ib*ns Average E /mK Average Ek /mK Average Ep /mK walkers

-------- ------------ --- ------------ --- ------------ --- --------

  2000   -89.4010789        736.581586       -825.982664          4988
  4000   -86.8499373        707.358765       -794.208703          4952
  6000   -87.3911178        680.584591       -767.975709          4968

Answer 1

If you have a CSV file, then I'd recommend using pandas .

In general though the solution is something like:

DELIMITER = ' '
data = []
with open('test.txt') as fr:
  for line in fr:
    first_col = line.split('{}'.format(DELIMITER))[0]
    data.append(first_col)

You can replace DELIMITER with whatever you want (spaces, tabs, etc.)

Or, as @cricket_007 says, use the Python csv module. Something like:

lines = csv.reader(open('test.txt', 'rb'), delimiter=DELIMITER)

Answer 2

the following code works if the separator between columns is a tab. Otherwise adjust to whatever the separator is.

import pandas as pd
df = pd.read_csv('sample.txt', header=None, usecols=[1], sep='\t')

note that despite read_csv claiming it reads csv's, it can read any text file with a consistent separator

Answer 3

The built-in csv module provides this functionality:

import csv
with open('data.dat', 'r') as csvfile:
     reader = csv.reader(csvfile, delimiter=' ')
     first_column = [ row[0] for row in reader ]

This code extracts the first column into a list as requested.