How to save each line of a file to a new file (every line a new file) and do that for multiple origi-CodePudding

I have 5 files from which i want to take each line (24 lines in total) and save it to a new file. I managed to find a code which will do that but they way it is, every time i have to manually change the number of the appropriate original file and of the file i want to save it to and also the number of each line every time.

The code:

x1= np.loadtxt("x_p2_40.txt")
x2= np.loadtxt("x_p4_40.txt")
x3= np.loadtxt("x_p6_40.txt")
x4= np.loadtxt("x_p8_40.txt")
x5= np.loadtxt("x_p1_40.txt")    

with open("x_p1_40.txt", "r") as file:
 content = file.read()
 first_line = content.split('\n', 1)[0]
with open("1_p_40_x.txt", "a" ) as f : 
       f.write("\n")
with open("1_p_40_x.txt", "a" ) as fa :     
       fa.write(first_line)
        
print(first_line)

I am a beginner at python, and i'm not sure how to make a loop for this, because i assume i need a loop?

Thank you!

CodePudding user response：

In order to read each of your input files, you can store them in a list and iterate over it with a for loop. Then we add every line to a single list with the function extend() :

inputFiles = ["x_p2_40.txt", "x_p4_40.txt", "x_p6_40.txt", "x_p8_40.txt", "x_p1_40.txt"]
outputFile = "outputfile.txt"

lines = []
for filename in inputFiles:
    with open(filename, 'r') as f:
        lines.extend(f.readlines())
        lines[-1]  = '\n'

Finally you can write all the line to your output file :

with open(outputFile, 'w') as f:
    f.write(''.join(lines))

CodePudding user response：

Since you have multiple files here, you could define their names in a list, and use a list comprehension to open file handles to them all:

input_files = ["x_p2_40.txt", "x_p4_40.txt", "x_p6_40.txt", "x_p8_40.txt", "x_p1_40.txt"]

file_handles = [open(f, "r") for f in input_files]

Since each of these file handles is an iterator that yields a single line every time you iterate over it, you could simply zip() all these file handles to iterate over them simultaneously. Also throw in an enumerate() to get the line numbers:

for line_num, files_lines in enumerate(zip(*file_handles), 1):
    out_file = f"{line_num}_p_40.txt"
    # Remove trailing whitespace on all lines, then add a newline
    files_lines = [f.rstrip()   "\n" for f in files_lines] 
    with open(out_file, "w") as of:
        of.writelines(files_lines)

With three files:

x_p2_40.txt:
2_1
2_2
2_3
2_4

x_p4_40.txt:
4_1
4_2
4_3
4_4

x_p6_40.txt:
6_1
6_2
6_3
6_4

I get the following output:

1_p_40.txt:
2_1
4_1
6_1

2_p_40.txt:
2_2
4_2
6_2

3_p_40.txt:
2_3
4_3
6_3

4_p_40.txt:
2_4
4_4
6_4

Finally, since we didn't use a context manager to open the original file handles, remember to close them after we're done:

for fh in file_handles:
    fh.close()

If you have files with an unequal number of lines and you want to create files for all lines, consider using itertools.zip_longest() instead of zip()