使用 Python 将字符串拆分为整数列表

Question

This method inputs a file and the directory of the file. It contains a matrix of data, and needs to copy the first 20 columns of each row after the given row name and the corresponding letter for the row. The first 3 lines of each file is skipped because it has unimportant information that is not needed, and it also doesn't need the data at the bottom of the file.

For example a file would look like:

unimportant information--------
 unimportant information--------
 -blank line
1 F -1 2 -3 4 5 6 7 (more columns of ints)
2 L 3 -1 3 4 0 -2 1 (more columns of ints)
3 A 3 -1 3 6 0 -2 5 (more columns of ints)
-blank line
unimportant information--------
unimportant information--------

The output of the method needs to print out a "matrix" in some given form.

So far the output gives a list of each row as a string, however I'm trying to figure out the best way to approach the problem. I don't know how to ignore the unimportant information at the end of the files. I don't know how to only retrieve the first 20 columns after the letter in each row, and I don't know how to ignore the row number and the row letter.

def pssmMatrix(self,ipFileName,directory):
    dir = directory
    filename = ipFileName
    my_lst = []

    #takes every file in fasta folder and put in files list
    for f in os.listdir(dir):
        #splits the file name into file name and its extension
        file, file_ext = os.path.splitext(f)

        if file == ipFileName:
            with open(os.path.join(dir,f)) as file_object:

                for _ in range(3):
                    next(file_object)
                for line in file_object:
                        my_lst.append(' '.join(line.strip().split()))
    return my_lst

Expected results:

['-1 2 -3 4 5 6 7'], ['3 -1 3 4 0 -2 1'], ['3 -1 3 6 0 -2 5']

Actual results:

['1 F -1 2 -3 4 5 6 7'], ['2 L 3 -1 3 4 0 -2 1'], ['3 A 3 -1 3 6 0 -2 5'],  [' '], [' unimportant info'], ['unimportant info']  

Answer 1

Try this solution.

    import re
    reg = re.compile(r'(?<=[0-9]\s[A-Z]\s)[0-9\-\s]+')

    text = """
    unimportant information--------

    unimportant information--------
    -blank line

    1 F -1 2 -3 4 5 6 7 (more columns of ints)

    2 L 3 -1 3 4 0 -2 1 (more columns of ints)

    3 A 3 -1 3 6 0 -2 5 (more columns of ints)"""

    ignore_start = 5  # 0,1,2,3 =  4
    expected_array = []
    for index, line in enumerate(text.splitlines()):
    if(index >= ignore_start):
            if reg.search(line):
            result = reg.search(line).group(0).strip()
            # Use Result
            expected_array.append(' '.join(result))

    print(expected_array)
    # Result: [
    #'- 1   2   - 3   4   5   6   7', 
    #'3   - 1   3   4   0   - 2   1', 
    #'3   - 1   3   6   0   - 2   5'
    #]

Answer 2

To drop the first two columns, you can change:

my_lst.append(' '.join(line.strip().split()))

to

my_lst.append(' '.join(line.strip().split()[2:]))

That will drop the first two columns after they've been split and before they've been joined back together.

To drop the last 3 irrelevant lines, maybe the simplest solution is just to change:

return my_lst

to

return my_lst[:-3]

That will return everything except the last 3 lines.

Answer 3

Ok so it looks to me like you have a file with certain lines that you want and the lines that you want always start with a number followed by a letter. So what we can do is apply a regular expression to this that only gets lines that match that pattern and only get the numbers after the pattern

The expression for this would look like (?<=[0-9]\\s[AZ]\\s)[0-9\\-\\s]+

import re

reg = re.compile(r'(?<=[0-9]\s[A-Z]\s)[0-9\-\s]+')

for line in file:
    if reg.search(line):
        result = reg.search(test).group(0)
        # Use Result
        my_lst.append(' '.join(result))

Hope that helps