Python：打印函数文本的自定义打印函数

Question

Python脚本：text = "abcde"print("text[::-1] : ", text[::-1])print("text[5:0:-1] : ", text[5:0:-1])输出：text[::-1] :  edcbatext[5:0:-1] :  edcb可以定义一个自定义函数来避免输入重复吗？例如：text = "abcde"def fuc(x):    print(x, ":", x)    fuc(text[::-1])fuc(text[5:0:-1])要求。输出：text[::-1] :  edcbatext[5:0:-1] :  edcb

Answer 1

在 python 3.8+ 中，您可以使用自记录表达式

>>> print(f"{a[::-1]=}") 

a[::-1]='edcba'

Answer 2

使用 f 字符串可以实现涉及字符串插值的解决方案。但是，f 字符串仅适用于 Python 3.8+

但是，我们可以很容易地实现字符串插值，如如何在 Python 中实现字符串插值

Current Modification 扩展了上述参考，以允许除了变量查找之外的表达式。

import sys

from re import sub

def interp(s):

  '''Implement simple string interpolation to handle

    "{var}" replaces var with its value from locals

    "{var=}" replaces var= with var=value, where value is the value of var from locals

    "{expressions} use eval to evaluate expressions (make eval safe by only allowing items from local functions and variables in expression'''

  # 1. Implement self-documenting expressions similar to https://docs.python.org/3/whatsnew/3.8.html#f-strings-support-for-self-documenting-expressions-and-debugging

  s1 = sub( r'{\s*(.*?)=\s*}', lambda m: m.group(1) + '=' + '{' + m.group(1) + '}', s)

  # Get the locals from the previous frame

  previous_frame = sys._getframe(1)  # i.e. current frame(0), previous is frame(1)

  d = previous_frame.f_locals        # get locals from previous frame

  # 2--Replace variable and expression with values

  # Use technique from http://lybniz2.sourceforge.net/safeeval.html to limit eval to make it safe

  # by only allowing locals from d and no globals

  s2 = sub(r'{\s*([^\s]+)\s*}', lambda m: str(d[m.group(1)]) if m.group(1) in d else str(eval(m.group(1), {}, d)), s1)

  return s2

# Test

a = "abcde"

print(interp("a has value {a}"))  # without self-doc =

#Output>>> a has value abcde

print(interp("{a[::-1]=}"))       # with self-doc =

#Output>>> a[::-1]=edcba

print(interp('{a[4:0:-1]=}'))     # with self-doc =

#Output>>> a[4:0:-1]=edcb

print(interp('sum {1+1}') # without self-doc =

#Output>>> sum 2

print(interp('{1+1=}'))  # with self-doc =

#Output>>> 1+1=2