Python - find pattern in file in max 4 lines of code-CodePudding

i have the following task. I have to find a specific pattern(word) in my file.txt(is a song centered on page) and to print out the row number the row which has the pattern in it getting rid of the left spaces. You can see the correct output here:


 92 Meant in croaking "Nevermore."
 99 She shall press, ah, nevermore!
107 Quoth the Raven, "Nevermore."
115 Quoth the Raven, "Nevermore."

and without this: my_str  = ' ' str(count)  ' '   line.lstrip(), it will print:

92 Meant in croaking "Nevermore."
99 She shall press, ah, nevermore!
107 Quoth the Raven, "Nevermore."
115 Quoth the Raven, "Nevermore."

This is my code, but i want to have only 4 lines of code
```python
def find_in_file(pattern,filename):

    my_str = ''
    with open(filename, 'r') as file:
        for count,line in enumerate(file):
            if pattern in line.lower():
                if count >= 10 and count <= 99:
                    my_str  = ' ' str(count)  ' '   line.lstrip()
                else:
                    my_str  = str(count)  ' '   line.lstrip()


    print(my_str)

CodePudding user response：

In fact, one line can be completed:

''.join(f' {count} {line.lstrip()}' if 10 <= count <= 99 else f'{count} {line.lstrip()}' for count, line in enumerate(file) if pattern in line.lower())

However, this seems a little too long...

According to the comment area, it can be simplified:

''.join(f'{count:3} {line.lstrip()}' for count, line in enumerate(file) if pattern in line.lower())

CodePudding user response：

def find_in_file(pattern,filename):
    with open(filename, 'r') as file:
        # 0 based line numbering, for 1 based use enumerate(file,1)
        for count,line in enumerate(file):
            if pattern in line.lower():
                print(f"{count:>3} {line.strip()}")

would be 4 lines of code (inside the function) and should be equivalent to what you got.

Possible in one line as well:

def find_in_file(pattern,filename):
    # 1 based line numbering
    return '\n'.join(f'{count:>3} {line.strip()}' for count, line in enumerate(file,1) if pattern in line.lower())

See pythons mini format language.

CodePudding user response：

You can use formatted strings to make sure the numbers always use three characters, even when they have only 1 or 2 digits.

I also prefer to use str.strip rather than str.lstrip, to get rid of trailing whitespace; in particular, lines read from the file will typically end with a linebreak, and then print will add a second linebreak, and we end up with too many linebreaks if we don't strip them away.

def find_in_file(pattern,filename):
    with open(filename, 'r') as file:
        for count,line in enumerate(file):
            if pattern in line.lower():
                print('{:3d} {}'.format(count, line.strip()))

find_in_file('nevermore','theraven.txt')
#  55 Quoth the Raven "Nevermore."
#  62 With such name as "Nevermore."
#  69 Then the bird said "Nevermore."
#  76 Of 'Never—nevermore'."
#  83 Meant in croaking "Nevermore."
#  90 She shall press, ah, nevermore!
#  97 Quoth the Raven "Nevermore."
# 104 Quoth the Raven "Nevermore."
# 111 Quoth the Raven "Nevermore."
# 118 Quoth the Raven "Nevermore."
# 125 Shall be lifted—nevermore!