搜索文本文件并获取python中不包含##的所有行

Question

I am trying to write a python script to read in a large text file from some modeling results, grab the useful data and save it as a new array. The text file is output in a way that has a ## starting each line that is not useful. I need a way to search through and grab all the lines that do not include the ##. I am used to using grep -v in this situation and piping to a file. I want to do it in python!

Thanks a lot.

-Tyler

Answer 1

I would use something like this:

fh = open(r"C:\Path\To\File.txt", "r")

raw_text = fh.readlines()
clean_text = []

for line in raw_text:
    if not line.startswith("##"):
        clean_text.append(line)

Or you could also clean the newline and carriage return non-printing characters at the same time with a small modification:

for line in raw_text:
    if not line.startswith("##"):
        clean_text.append(line.rstrip("\r\n"))

You would be left with a list object that contains one line of required text per element. You could split this into individual words using string.split() which would give you a nested list per original list element which you could easily index (assuming your text has whitespaces of course).

clean_text[4][7]

would return the 5th line, 8th word.

Hope this helps.

[Edit: corrected indentation in loop]

Answer 2

My suggestion would be to do the following:

listoflines = [ ] 

with open(.txt, "r") as f:     # .txt = file, "r" = read
    for line in f:
        if line[:2] != "##": #Read until the second character 
            listoflines.append(line)


print listoflines

If you're feeling brave, you can also do the following, CREDITS GO TO ALEX THORNTON:

listoflines = [l for l in f if not l.startswith('##')]

The other answer is great as well, especially teaching the .startswith function, but I think this is the more pythonic way and also has the advantage of automatically closing the file as soon as you're done with it.