human
Solve the following python programming problem given the following code:
def _format_source_error(filename, context, lineno) : 
    ' A helper function which generates an error string.

    This function handles the work of reading the lines of the file
    which bracket the error, and formatting a string which points to
    the offending line. The output is similar to:

    File "foo.py", line 42, in bar()
           41 def bar():
    ---->  42     a = a + 1
           43     return a

    Parameters
    ----------
    filename : string
        The name of the offending file.

    context : string
        The string name of the context scope in which the error
        occured. In the sample above, the context is \'bar\'.

    lineno : int
        The integer line number of the offending line.

    Returns
    -------
    result : string
        A nicely formatted string for including in an exception. If the
        file cannot be opened, the source lines will note be included.

    ' 
    text = ('File "%s", line %d, in %s()' % (filename, lineno, context)) 
    start_lineno = max (0, (lineno - 1)) 
    end_lineno = (start_lineno + 2) 
    lines = [] 
    try : 
        with open (filename, 'r') as f : 
            for (idx, line) in enumerate (f, 1) : 
                if ((idx >= start_lineno) and (idx <= end_lineno)) : 
                    lines.append ((idx, lno)) 
else : 
                    if (idx > end_lineno) : 
                        break 
except IOError : 
        pass 
if (len (lines) > 0) : 
        digits = str (len (str (end_lineno))) 
        line_templ = (('
----> %' + digits) + 'd %s') 
        other_templ = (('
      %' + digits) + 'd %s') 
        for (lno, line) in lines : 
            line = line.rstrip () 
            if (lno == lineno) : 
                text += (line_templ % (lno, line)) 
else : 
                text += (other_templ % (lno, line)) 
return text 

    
    

    Fix the buggy line: lines.append ((idx, lno))
assistant
#Here is Python code to fix the buggy line: lines.

lines.append ((idx, line))