使用 Python 替换文本文件中一行中的一列

Question

I have a text file and it has the following contents

    #                                       
    #   Keywords:                                   
    #                                                                           
    LiFePO4                                     
    end                                     
    name    Li5FeO4                                 
    cell                                        
        18.557309   18.316802   9.125725    90.047539   90.100646   90.060551 0 0 0 0 0 0           
    fractional  1                                   
    Li  core    0.06001     0.059408    0.849507    1   1   0   0   0   0
    Li1 core    0.025416    0.339078    0.128746    1   1   0   0   0   0
    Li2 core    0.02517     0.838929    0.130747    1   1   0   0   0   0
    Li3 core    0.525498    0.339179    0.127632    1   1   0   0   0   0
    Li4 core    0.524753    0.841333    0.129329    1   1   0   0   0   0
    Li5 core    0.179907    0.158182    0.634012    1   1   0   0   0   0
    Li6 core    0.180817    0.666028    0.628327    1   1   0   0   0   0

This is the input that I need to supply to a tool which used in some research application. Now I need to replace the 0 on the first line which starts with Li on the third column from the last. That is, there are four zeros towards the end in each of the lines starting with Li . I need to replace the second zero and so the file will have the the contents as follows:

    #                                       
    #   Keywords:                                   
    #                                                                           
    LiFePO4                                     
    end                                     
    name    Li5FeO4                                 
    cell                                        
        18.557309   18.316802   9.125725    90.047539   90.100646   90.060551 0 0 0 0 0 0           
    fractional  1                                   
    Li  core    0.06001     0.059408    0.849507    1   1   0   1   0   0
    Li1 core    0.025416    0.339078    0.128746    1   1   0   0   0   0
    Li2 core    0.02517     0.838929    0.130747    1   1   0   0   0   0
    Li3 core    0.525498    0.339179    0.127632    1   1   0   0   0   0
    Li4 core    0.524753    0.841333    0.129329    1   1   0   0   0   0
    Li5 core    0.179907    0.158182    0.634012    1   1   0   0   0   0
    Li6 core    0.180817    0.666028    0.628327    1   1   0   0   0   0

This has to be done a number of times for the zeros in various positions and I have the following code. There are some more operations that I am doing in the same code.

import os
import shutil
import time


def edit_file(column, next_column):
    # Processing x.gin file
    file_name = './' + column + '.gin'
    file_handler = open(file_name, 'r')
    print("Processing " + file_name + " file")
    contents = file_handler.readlines()
    find_line = contents[14]
    find_line_array = find_line.split('\t')

    print(find_line_array)

    # change values according to the file name
    if column == 'x':
        find_line_array[8] = 1
    elif column == 'y':
        print(contents)
        print(find_line_array)
        find_line_array[9] = 1
    elif column == 'z':
        find_line_array[10] = 1
    elif column == 'xy':
        find_line_array[8] = 1
        find_line_array[9] = 1
    elif column == 'yz':
        find_line_array[9] = 1
        find_line_array[10] = 1
    elif column == 'xz':
        find_line_array[8] = 1
        find_line_array[10] = 1

    formatted = '\t'.join(map(str, find_line_array))
    contents[14] = formatted
    with open(file_name, 'w') as f:
        for item in contents:
            f.write("%s\n" % item)
    print("Formatting completed for " + file_name)

    print('Executing GULP command ----')

    gulp_command = 'gulp ' + column + '.gin > ' + column + '.gout'
    print(gulp_command)
    shutil.copy(file_name, next_column+'.gin')

    file_handler.close()
    os.system(gulp_command)

    while not os.path.exists('./Li.grs'):
        print('Waiting for output file')
        time.sleep(1)

    if os.path.isfile('./Li.grs'):
        print('renaming file')
        os.rename('./Li.grs', next_column+'.gin')
        os.rename('./Li.grs', column+'.grs')

    return True


if __name__ == '__main__':
    print('Starting Execution')
    column_list = ['x', 'y', 'xy', 'yz', 'xz']
    print(column_list)
    for index, column in enumerate(column_list):
        if column != 'xz':
            edit_file(column, column_list[index + 1])
        else:
            edit_file(column, 'xz')
    print('Execution completed')

I am replacing it correctly and rewriting the file. But this file doesn't appears to be in correct format as it has additional new lines. Is it possible that I can rewrite the single line only and so that I can keep the file in exact same format.

Answer 1

i created a function for that. try this

def replace(filename,row,column,value):
    columnspan = "     "
    data = open(filename).read().split("\n")
    for i in range(len(data)):
        data[i] = data[i].split(columnspan)
    data[row][column] = value
    write=""
    for i in range(len(data)):
        for x in range(len(data[i])):
            write+=(str(data[i][x])+columnspan)
        write += "\n"
    write.strip()
    file = open(filename,"w")
    file.write(write)
    file.close()

Answer 2

You can use regex to find and update text:

import re

with open('input.txt', 'r') as f1, open('output.txt', 'w') as f2:
    data = f1.read()
    match = re.findall('Li\s+\w+\s+\d+\.\d+\s+\d\.\d+\s+\d\.\d+\s+\d\s+\d\s+\d\s+\d', data)
    for m in match:
        data = data.replace(m, f'{m[:-1]}1')
    f2.write(data)

Output:

#
#   Keywords:
#
LiFePO4
end
name    Li5FeO4
cell
    18.557309   18.316802   9.125725    90.047539   90.100646   90.060551 0 0 0 0 0 0
fractional  1
Li  core    0.06001     0.059408    0.849507    1   1   0   1   0   0
Li1 core    0.025416    0.339078    0.128746    1   1   0   0   0   0
Li2 core    0.02517     0.838929    0.130747    1   1   0   0   0   0
Li3 core    0.525498    0.339179    0.127632    1   1   0   0   0   0
Li4 core    0.524753    0.841333    0.129329    1   1   0   0   0   0
Li5 core    0.179907    0.158182    0.634012    1   1   0   0   0   0
Li6 core    0.180817    0.666028    0.628327    1   1   0   0   0   0