我这里有实现静态变量装饰器的代码。但是,我发现如果我多次运行这个函数,每次调用该函数时都不会重新初始化静态变量。
def static_vars(**kwargs):
def decorate(func):
for k in kwargs:
setattr(func, k, kwargs[k])
return func
return decorate
@static_vars(count=0)
def rolling_serial(val):
'''
For a vector V = [v_1, ..., V_N] returns a serial
index.
so for V = [1, 1, 1, 3, 1, 1, 1]
a resulting vector will be generated
V_hat = [1, 2, 3, 4, 5, 6, 7]
'''
temp = rolling_serial.count
rolling_serial.count += 1
return temp
# invoke it like this
from useful import (rolling_serial)
df = <...some dataframe with a column called ts>
self.df['ts_index'] = self.df.ts.apply(rolling_serial)
# Example output a new column, sa: [1, 2, 3, ..., N]
# My issue arises if I run it again
df = <...some dataframe with a column called ts>
self.df['ts_index'] = self.df.ts.apply(rolling_serial)
# output: [N+1, N+2, ...] instead of restarting at 0
如果我重新启动jupyter内核,静态变量就会清除。但我不希望重启内核。有谁可以帮我?
发布于 2018-07-20 01:52:25
你的装饰器只被调用一次,而不是每次调用你的函数时。确实,它是在定义时调用的:
def static_vars(**kwargs):
def decorate(func):
for k in kwargs:
print(kwargs)
setattr(func, k, kwargs[k])
return func
return decorate
@static_vars(count=0)
def rolling_serial(val):
'''
For a vector V = [v_1, ..., V_N] returns a serial
index.
so for V = [1, 1, 1, 3, 1, 1, 1]
a resulting vector will be generated
V_hat = [1, 2, 3, 4, 5, 6, 7]
'''
temp = rolling_serial.count
rolling_serial.count += 1
return temp
print('---- BEGIN ----')
print(rolling_serial(10))
print(rolling_serial(20))
print(rolling_serial(30))
打印:
{'count': 0}
---- BEGIN ----
0
1
2
您在static_vars()
中作为参数的kwargs
将成为闭包,并且会在每次调用rolling_serial()
时递增。
一种解决方案是通过globals()传递变量:
# This function creates decorator:
def static_vars(**global_kwargs):
# This is decorator:
def decorate(func):
# This function is called every time:
def _f(*args, **kwargs):
for k in global_kwargs:
globals()[func.__name__+'_'+k] = global_kwargs[k]
return func(*args, **kwargs)
return _f
return decorate
@static_vars(count=0, temp=40)
def rolling_serial():
global rolling_serial_count, rolling_serial_temp
temp1, temp2 = rolling_serial_count, rolling_serial_temp
rolling_serial_count += 1
rolling_serial_temp += 1
return temp1, temp2
print(rolling_serial()) # prints (0, 40)
print(rolling_serial()) # prints (0, 40)
print(rolling_serial()) # prints (0, 40)
发布于 2018-07-20 09:56:38
@
装饰器是一个障碍,因为它的目的是在函数定义时进行一次装饰。
因此,对传入函数进行裁剪并将其简化为setattr
关键字args。这是您必须手动完成的工作,去掉了@
语法速记。
def static_vars(func, **kwargs):
for k in kwargs:
setattr(func, k, kwargs[k])
return func
def rolling_serial(val):
temp = rolling_serial.count
rolling_serial.count += 1
return temp
static_vars(rolling_serial, count=0)
print (rolling_serial(3))
print (rolling_serial(3))
#reset it
static_vars(rolling_serial, count=0)
print (rolling_serial(3))
输出:
0
1
0
另外,FWIW,你不使用val
,而且dataframe不是紧要关头的,最好只发布rolling_serial
的一些预期结果。
https://stackoverflow.com/questions/51428879
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