一行Python代码可以知道其缩进嵌套级别吗？

From something like this:

print(get_indentation_level())

    print(get_indentation_level())

        print(get_indentation_level())

I would like to get something like this:

1
2
3

Can the code read itself in this way?

All I want is the output from the more nested parts of the code to be more nested. In the same way that this makes code easier to read, it would make the output easier to read.

Of course I could implement this manually, using e.g. .format(), but what I had in mind was a custom print function which would print(i*' ' + string) where i is the indentation level. This would be a quick way to make readable output on my terminal.

Is there a better way to do this which avoids painstaking manual formatting?

If you want indentation in terms of nesting level rather than spaces and tabs, things get tricky. For example, in the following code:

if True:
    print(
get_nesting_level())

the call to get_nesting_level is actually nested one level deep, despite the fact that there is no leading whitespace on the line of the get_nesting_level call. Meanwhile, in the following code:

print(1,
      2,
      get_nesting_level())

the call to get_nesting_level is nested zero levels deep, despite the presence of leading whitespace on its line.

In the following code:

if True:
  if True:
    print(get_nesting_level())

if True:
    print(get_nesting_level())

the two calls to get_nesting_level are at different nesting levels, despite the fact that the leading whitespace is identical.

In the following code:

if True: print(get_nesting_level())

is that nested zero levels, or one? In terms of INDENT and DEDENT tokens in the formal grammar, it’s zero levels deep, but you might not feel the same way.

If you want to do this, you’re going to have to tokenize the whole file up to the point of the call and count INDENT and DEDENT tokens. The tokenize module would be very useful for such a function:

import inspect
import tokenize

def get_nesting_level():
    caller_frame = inspect.currentframe().f_back
    filename, caller_lineno, _, _, _ = inspect.getframeinfo(caller_frame)
    with open(filename) as f:
        indentation_level = 0
        for token_record in tokenize.generate_tokens(f.readline):
            token_type, _, (token_lineno, _), _, _ = token_record
            if token_lineno > caller_lineno:
                break
            elif token_type == tokenize.INDENT:
                indentation_level += 1
            elif token_type == tokenize.DEDENT:
                indentation_level -= 1
        return indentation_level

Yeah, that’s definitely possible, here’s a working example:

import inspect

def get_indentation_level():
    callerframerecord = inspect.stack()[1]
    frame = callerframerecord[0]
    info = inspect.getframeinfo(frame)
    cc = info.code_context[0]
    return len(cc) - len(cc.lstrip())

if 1:
    print get_indentation_level()
    if 1:
        print get_indentation_level()
        if 1:
            print get_indentation_level()

You can use sys.current_frame.f_lineno in order to get the line number. Then in order to find the number of indentation level you need to find the previous line with zero indentation then be subtracting the current line number from that line’s number you’ll get the number of indentation:

import sys
current_frame = sys._getframe(0)

def get_ind_num():
    with open(__file__) as f:
        lines = f.readlines()
    current_line_no = current_frame.f_lineno
    to_current = lines[:current_line_no]
    previous_zoro_ind = len(to_current) - next(i for i, line in enumerate(to_current[::-1]) if not line[0].isspace())
    return current_line_no - previous_zoro_ind

Demo:

if True:
    print get_ind_num()
    if True:
        print(get_ind_num())
        if True:
            print(get_ind_num())
            if True: print(get_ind_num())
# Output
1
3
5
6

If you want the number of the indentation level based on the previouse lines with : you can just do it with a little change:

def get_ind_num():
    with open(__file__) as f:
        lines = f.readlines()

    current_line_no = current_frame.f_lineno
    to_current = lines[:current_line_no]
    previous_zoro_ind = len(to_current) - next(i for i, line in enumerate(to_current[::-1]) if not line[0].isspace())
    return sum(1 for line in lines[previous_zoro_ind-1:current_line_no] if line.strip().endswith(':'))

Demo:

if True:
    print get_ind_num()
    if True:
        print(get_ind_num())
        if True:
            print(get_ind_num())
            if True: print(get_ind_num())
# Output
1
2
3
3

And as an alternative answer here is a function for getting the number of indentation (whitespace):

import sys
from itertools import takewhile
current_frame = sys._getframe(0)

def get_ind_num():
    with open(__file__) as f:
        lines = f.readlines()
    return sum(1 for _ in takewhile(str.isspace, lines[current_frame.f_lineno - 1]))

To solve the ”real” problem that lead to your question you could implement a contextmanager which keeps track of the indention level and make the with block structure in the code correspond to the indentation levels of the output. This way the code indentation still reflects the output indentation without coupling both too much. It is still possible to refactor the code into different functions and have other indentations based on code structure not messing with the output indentation.

#!/usr/bin/env python
# coding: utf8
from __future__ import absolute_import, division, print_function


class IndentedPrinter(object):

    def __init__(self, level=0, indent_with='  '):
        self.level = level
        self.indent_with = indent_with

    def __enter__(self):
        self.level += 1
        return self

    def __exit__(self, *_args):
        self.level -= 1

    def print(self, arg='', *args, **kwargs):
        print(self.indent_with * self.level + str(arg), *args, **kwargs)


def main():
    indented = IndentedPrinter()
    indented.print(indented.level)
    with indented:
        indented.print(indented.level)
        with indented:
            indented.print('Hallo', indented.level)
            with indented:
                indented.print(indented.level)
            indented.print('and back one level', indented.level)


if __name__ == '__main__':
    main()

Output:

0
  1
    Hallo 2
      3
    and back one level 2

>>> import inspect
>>> help(inspect.indentsize)
Help on function indentsize in module inspect:

indentsize(line)
    Return the indent size, in spaces, at the start of a line of text.