Python，如何创建仅包含以特定短语开头的特定行的文件

Question

I have certain lines of data within a large file that the only lines I need are the ones that begin with "ABC". I wasn't sure if there was a way to pull all these lines and create a new folder to contain only the xyz coordinates within it, as I need to find the standard deviation and think that would be the best way to go about It. (An example line of the overall file that's being worked with: "ABC 3.2650481094164707 5.5229511970397915 4.8978200015718540")

Ideally it would be a file with column format so that the example coordinates are as such: "3.2650481094164707, 5.5229511970397915, 4.8978200015718540" so that there is an x column, ay column and az column

The following is the format of the code to obtain just the ABC lines, but I'm unsure how to make them into their own file

with open("file",'r') as f:
    sums = [0] * 3
    points = 0
    for line in f:
        if line.startswith("ABC "):
            points += 1
            for i, x in enumerate(line.split()[1:]):
                sums[i] += float(x)

Answer 1

To get lines from the input file which starts with ABC and write it to CSV file you can use next example:

import csv

abc_lines = []
with open("your_file.txt", "r") as f_in:
    for line in map(str.strip, f_in):
        if not line.startswith("ABC "):
            continue
        abc_lines.append(line.split()[1:])


with open("data.csv", "w") as f_out:
    writer = csv.writer(f_out)

    writer.writerow(["X", "Y", "Z"])
    writer.writerows(abc_lines)

Creates data.csv :

X,Y,Z
3.2650481094164707,5.5229511970397915,4.8978200015718540
4.2650481094164707,6.5229511970397915,7.8978200015718540

---

Input file:

ABC 3.2650481094164707 5.5229511970397915 4.8978200015718540
XYZ 3.2650481094164707 5.5229511970397915 4.8978200015718540
ABC 4.2650481094164707 6.5229511970397915 7.8978200015718540