Python - 如何获取特定范围内变量的字典（甚至列表）；比 locals/globals() 更具体
Python - How to obtain a dictionary(or even a list) of variables in a specific scope; more specific than locals/globals()
所以，标题几乎说明了一切。
例如让我们看下面的示例代码：
## How can I obtain a dict/list (like locals()) of all the variables in second and/or third layer scopes via a command
# coming from the first layer?
## Or another example would be how could I obtain the variables "locals() style" from the thirdlayer via a
# command from the second layer?
# essentially can a parent function/class access a list/dict of a child function
# or class??
def firstLayer():
a = 4.7
q = locals()
print(q)
# local vars of 1st layer
def secondlayer():
b = 7
r = locals()
print(r)
# local vars of 2nd layer
def thirdlayer():
c = False
s = locals()
i = globals()
print('c:\n', c, "\nglobals from 3rd layer:\n\t", i)
# local vars of 3rd layer
thirdlayer()
secondlayer()
firstLayer()
sample_var = globals()
print(sample_var)
# returns the list of global variables
重申我在代码的注释中所说的内容，本质上是他们可以通过任何方式获得 'child' 范围内所有局部变量的列表？我知道功能已关闭，但如果他们无法做到这一点，那么他们可以使用更复杂的代码来实现这一点，我可以将其集成到一个函数中，或者在必要时 class。
编辑：
进一步阐述；这是我现在的情况。
def varsfunc():
font1 = "Harlow Solid"
grey = '#454545'
font2 = 'Mistral'
font3 = 'Italic 35px Times New Roman'
pnk = 'pink'
grn = 'green'
return locals()
本质上，我正在创建一个模块，用户必须创建某种类型的函数，他们列出他们想要声明的所有变量以用于修改 css 文件。本质上，我想让用户不必键入 "return locals()"。我想通过让 end-users 将上面的示例函数包装在一个装饰器中来实现它，该装饰器相当于返回我想要的确切范围的 locals() 。装饰器对我不起作用，因为它在外部范围内。
更清楚：
我需要一个 decorator/function 来包装另一个函数（即装饰器），它可以访问和创建一个 child 元素的列表。
def module_decorator_func_thing():
r = command_that_acts_like_locals()_but_for_child_scopes
def user_var_list():
font1 = 'green'
font2 = 'pink'
# back in "module_decorator_func_thing"'s scope
print(r) # this variable should contain only a dict/list containing the
# the following:
# r = {'font1': 'green', 'font2': 'pink')
目前用户需要这样做：
def vars_func_container():
font1 = 'green'
font2 = 'pink'
return locals() # <---- I want the user to not have to type this and for
# a function decorator to take care of it instead possibly.
@aguy 和其他希望了解更多信息的人的信息。
我通过你们的提示获得的 dictionary/list 将被发送到这个函数来完成程序的真正工作。
（如果我要开始使用列表，我需要转换成字典，但这对我来说没问题。）
变量字典与此函数一起使用 "compile/compyle"（双关语 'Python' + '编译）并插入 "variables" 参数中。例如你像这样执行函数。
compyle("My sample title", return_stylesheet_from_func(*insert .css filename),
return_variables_from_function(*insert function containing variables*), "**True/False to turn on compilation**",
"**True/False to turn on annotations/suggestions**")
def compyle(title, style_sheet, variables, boolean=False, boolean2=True):
"""
:param title: The name you wish your .css file to be named.
:param style_sheet: The name of the multi-line string that will compose your .css file
:param variables: The name of the dictionary containing your .pcss variables
:param boolean: A.K.A the "Compiler Parameter" - Turns the compiler on or off
:param boolean2: A.K.A the "Annotation Parameter" - Turns annotations on or off
:return: returns compiled .pcss text as normal .css style text to be utilized with .html
"""
# -----------------------------------
file_name = title + ".css"
replace_num = len(variables.keys())
counter = replace_num
content = style_sheet
# -----------------------------------
# add theme support with namedtuple's formatted to mimic structs in C/C++
# this will be a major feature update as well as a nice way to allow the future prospect of integrating C/C++ into
# the compiler. Info:
for k, v in variables.items():
counter -= 1
content = content.replace(k, v, replace_num)
if counter == 0:
break
else:
pass
looped_content = str(content)
id_content = looped_content.replace("hash_", "#")
output = id_content.replace("dot_", ".")
if boolean is True:
if boolean2 is True:
output = " /* --- Pyle Sheet --- */\n" + output
with open(file_name, 'w') as writ:
writ.write(output)
writ.close()
print('compiled successfully; The file was saved as ' + "\"" + file_name + "\".")
elif boolean2 is False:
pass
else:
logging.warning("An Error Occurred - see module, documentation, or online Q&A for assistance.")
elif boolean is False:
if boolean2 is True:
print('compiled successfully; The file ' + "\"" + file_name + "\"" + "was not saved/created.")
elif boolean2 is False:
pass
else:
logging.warning("An Error Occurred - see module, documentation, or online Q&A for assistance.")
else:
logging.warning('An Error Occurred with the Compile Parameter (See: boolean in pyle_sheets source file) - \ '
'see module, documentation, or online Q&A for assistance.')
如果不深入研究，我看不出有什么方法可以做到这一点；以下是我想出的最简单的解决方案。
工作原理
使用 ast 模块，我们遍历给定函数的代码并找到所有赋值。这些在给定的命名空间中进行评估，并返回此命名空间。
密码
import ast
import functools
import inspect
def returnAssignments(f):
@functools.wraps(f)
def returner():
assignments = dict()
for node in ast.walk(ast.parse(inspect.getsource(f))):
if isinstance(node, ast.Assign):
exec(compile(ast.Module([node]), '<ast>', 'exec'),
globals(),
assignments)
return assignments
return returner
用法
from ra import returnAssignments
@returnAssignments
def foo():
this = 'something'
that = 37
the_other = object()
print(foo())
输出
rat@pandion:~/tmp$ python test.py
{'this': 'something', 'that': 37, 'the_other': <object object at 0x10205b130>}
我想知道我在这里提供的这种粗略的解决方案是否对您有用。请注意，我还没有在所有情况下对其进行测试，因此它可能有点粗糙。此外，它 returns 一切都作为字符串，您可能希望进一步更改这种行为。
定义函数：
def get_local_vars_from_function(f):
import inspect
s = inspect.getsourcelines(f)[0]
d = {}
for l in s:
if '=' in l:
var, val = l.split('=')
var = var.strip()
val = val.strip()
d[var] = val
return d
然后使用它：
In[91]: get_local_vars_from_function(user_var_list)
Out[91]: {'font1': "'green'", 'font2': "'pink'"}
所以，标题几乎说明了一切。
例如让我们看下面的示例代码：
## How can I obtain a dict/list (like locals()) of all the variables in second and/or third layer scopes via a command
# coming from the first layer?
## Or another example would be how could I obtain the variables "locals() style" from the thirdlayer via a
# command from the second layer?
# essentially can a parent function/class access a list/dict of a child function
# or class??
def firstLayer():
a = 4.7
q = locals()
print(q)
# local vars of 1st layer
def secondlayer():
b = 7
r = locals()
print(r)
# local vars of 2nd layer
def thirdlayer():
c = False
s = locals()
i = globals()
print('c:\n', c, "\nglobals from 3rd layer:\n\t", i)
# local vars of 3rd layer
thirdlayer()
secondlayer()
firstLayer()
sample_var = globals()
print(sample_var)
# returns the list of global variables
重申我在代码的注释中所说的内容，本质上是他们可以通过任何方式获得 'child' 范围内所有局部变量的列表？我知道功能已关闭，但如果他们无法做到这一点，那么他们可以使用更复杂的代码来实现这一点，我可以将其集成到一个函数中，或者在必要时 class。
编辑： 进一步阐述；这是我现在的情况。
def varsfunc():
font1 = "Harlow Solid"
grey = '#454545'
font2 = 'Mistral'
font3 = 'Italic 35px Times New Roman'
pnk = 'pink'
grn = 'green'
return locals()
本质上，我正在创建一个模块，用户必须创建某种类型的函数，他们列出他们想要声明的所有变量以用于修改 css 文件。本质上，我想让用户不必键入 "return locals()"。我想通过让 end-users 将上面的示例函数包装在一个装饰器中来实现它，该装饰器相当于返回我想要的确切范围的 locals() 。装饰器对我不起作用，因为它在外部范围内。
更清楚： 我需要一个 decorator/function 来包装另一个函数（即装饰器），它可以访问和创建一个 child 元素的列表。
def module_decorator_func_thing():
r = command_that_acts_like_locals()_but_for_child_scopes
def user_var_list():
font1 = 'green'
font2 = 'pink'
# back in "module_decorator_func_thing"'s scope
print(r) # this variable should contain only a dict/list containing the
# the following:
# r = {'font1': 'green', 'font2': 'pink')
目前用户需要这样做：
def vars_func_container():
font1 = 'green'
font2 = 'pink'
return locals() # <---- I want the user to not have to type this and for
# a function decorator to take care of it instead possibly.
@aguy 和其他希望了解更多信息的人的信息。 我通过你们的提示获得的 dictionary/list 将被发送到这个函数来完成程序的真正工作。 （如果我要开始使用列表，我需要转换成字典，但这对我来说没问题。） 变量字典与此函数一起使用 "compile/compyle"（双关语 'Python' + '编译）并插入 "variables" 参数中。例如你像这样执行函数。
compyle("My sample title", return_stylesheet_from_func(*insert .css filename),
return_variables_from_function(*insert function containing variables*), "**True/False to turn on compilation**",
"**True/False to turn on annotations/suggestions**")
def compyle(title, style_sheet, variables, boolean=False, boolean2=True):
"""
:param title: The name you wish your .css file to be named.
:param style_sheet: The name of the multi-line string that will compose your .css file
:param variables: The name of the dictionary containing your .pcss variables
:param boolean: A.K.A the "Compiler Parameter" - Turns the compiler on or off
:param boolean2: A.K.A the "Annotation Parameter" - Turns annotations on or off
:return: returns compiled .pcss text as normal .css style text to be utilized with .html
"""
# -----------------------------------
file_name = title + ".css"
replace_num = len(variables.keys())
counter = replace_num
content = style_sheet
# -----------------------------------
# add theme support with namedtuple's formatted to mimic structs in C/C++
# this will be a major feature update as well as a nice way to allow the future prospect of integrating C/C++ into
# the compiler. Info:
for k, v in variables.items():
counter -= 1
content = content.replace(k, v, replace_num)
if counter == 0:
break
else:
pass
looped_content = str(content)
id_content = looped_content.replace("hash_", "#")
output = id_content.replace("dot_", ".")
if boolean is True:
if boolean2 is True:
output = " /* --- Pyle Sheet --- */\n" + output
with open(file_name, 'w') as writ:
writ.write(output)
writ.close()
print('compiled successfully; The file was saved as ' + "\"" + file_name + "\".")
elif boolean2 is False:
pass
else:
logging.warning("An Error Occurred - see module, documentation, or online Q&A for assistance.")
elif boolean is False:
if boolean2 is True:
print('compiled successfully; The file ' + "\"" + file_name + "\"" + "was not saved/created.")
elif boolean2 is False:
pass
else:
logging.warning("An Error Occurred - see module, documentation, or online Q&A for assistance.")
else:
logging.warning('An Error Occurred with the Compile Parameter (See: boolean in pyle_sheets source file) - \ '
'see module, documentation, or online Q&A for assistance.')
如果不深入研究，我看不出有什么方法可以做到这一点；以下是我想出的最简单的解决方案。
工作原理
使用 ast 模块，我们遍历给定函数的代码并找到所有赋值。这些在给定的命名空间中进行评估，并返回此命名空间。
密码
import ast
import functools
import inspect
def returnAssignments(f):
@functools.wraps(f)
def returner():
assignments = dict()
for node in ast.walk(ast.parse(inspect.getsource(f))):
if isinstance(node, ast.Assign):
exec(compile(ast.Module([node]), '<ast>', 'exec'),
globals(),
assignments)
return assignments
return returner
用法
from ra import returnAssignments
@returnAssignments
def foo():
this = 'something'
that = 37
the_other = object()
print(foo())
输出
rat@pandion:~/tmp$ python test.py
{'this': 'something', 'that': 37, 'the_other': <object object at 0x10205b130>}
我想知道我在这里提供的这种粗略的解决方案是否对您有用。请注意，我还没有在所有情况下对其进行测试，因此它可能有点粗糙。此外，它 returns 一切都作为字符串，您可能希望进一步更改这种行为。
定义函数：
def get_local_vars_from_function(f):
import inspect
s = inspect.getsourcelines(f)[0]
d = {}
for l in s:
if '=' in l:
var, val = l.split('=')
var = var.strip()
val = val.strip()
d[var] = val
return d
然后使用它：
In[91]: get_local_vars_from_function(user_var_list)
Out[91]: {'font1': "'green'", 'font2': "'pink'"}