删除特定行python之前的所有行

Question

I have a file-

First Line
Second Line
Third Line
Fourth Line
Fifth Line
Sixth Line

Now I want to write the contents of the file after the fourth line. therefore the new file would be-

Fifth Line
Sixth Line

My attempt-

cnt=0
for root, dirs, files in os.walk(currentdir):
    for name in files:



        f = open(root+"/"+name,'r')
        lines = f.readlines()
        f.close()
        f = open(os.path.join(root, name),"w")
        for line in lines:
            cnt+=1 
            if line =="fourth Line"+"\n":
                break:
        f.seek(cnt)
        f.write(line)
        f.close()

Answer 1

for path, dirs, files in os.walk(currentdir):
    for name in files:
        with open(os.path.join(path, name)) as f:
            data = f.readlines()
        pos = data.index("Fourth Line\n")
        with open(os.path.join(path, name), 'w') as f:
            f.writelines(data[pos:])

Answer 2

You can just get the lines in a file, slice the array, and then write them back in:

with open('filename.txt', 'r+') as f:
    lines = f.readlines()
    f.write(lines[5:])

If you wanted to you could read and write in separate with blocks, or whatever else.

Answer 3

You can load the file lines to a list, and after split the list keeping only the lines you want and saving on the file. Example, if you want the content after the 4th line:

f = open("file.txt", 'r')
lines = f.readlines()
f.close()
lines = lines[4:]
f = open("file.txt", 'w')
f.write(lines)
f.close()

Answer 4

you could always place a counter in loop, and start writing the second file when counter reaches 4. Something like:

with open('a', 'w') as a, open('b', 'w') as b:
do_something()

or:

f1 = open(path, 'w')
f1.write(data)
f1.close()
counter += 1
if counter >= 4:
f2 = open(path, 'w')
f2.write(data)
f2.close()