字典理解产生看似没有根据的 NameError

Question

我正在使用 brian2 到运行神经网络模拟。为了在每次模拟期间记录数据，我正在创建 brian2 的 SpikeMonitor class 的几个实例。我想将这些监视器存储在使用字典理解创建的字典中。

作为测试，我在交互式会话中执行以下命令：

In [1]: import brian2

In [2]: pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)

In [3]: record_pops = ['pe_mt']

In [4]: {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in record_pops}
Out[4]: {'mon_pe_mt': <SpikeMonitor, recording spikemonitor>}

一切看起来都很棒。但是现在当我把这段代码移到下面的函数中时

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    return {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in
            record_pops}

并调用它，我得到以下错误

In [9]: tests.test_record()
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-9-4d3d585b2c97> in <module>()
----> 1 tests.test_record()

/home/daniel/Science/dopa_net/brian/ardid/tests.py in test_record()
     61     record_pops = ['pe_mt']
     62     return {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in
---> 63                 record_pops}
     64     # DEBUG ###################
     65     #monitors = utils.record(['pe_mt'], 'spikes', None, None, pe_mt, None, None)

/home/daniel/Science/dopa_net/brian/ardid/tests.py in <dictcomp>((pop,))
     60     # DEBUG ###################
     61     record_pops = ['pe_mt']
---> 62     return {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in
     63                 record_pops}
     64     # DEBUG ###################

/home/daniel/Science/dopa_net/brian/ardid/tests.py in <module>()

NameError: name 'pe_mt' is not defined

这是怎么回事？ 'pe_mt' 是在函数内定义的。

请注意，如果我将字典理解更改为列表理解，如

return [brian2.SpikeMonitor(eval(pop)) for pop in record_pops]

没有报错！我得到了 SpikeMonitor 个对象的列表，这些对象定义得当。

现在已删除的一个答案建议我使用 locals()[pop] 而不是 eval(pop)。请注意，这会引发等效错误：

In [20]: tests.test_record()
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-20-4d3d585b2c97> in <module>()
----> 1 tests.test_record()

/home/daniel/Science/dopa_net/brian/ardid/tests.py in test_record()
     61     record_pops = ['pe_mt']
     62     return {'mon_' + pop: brian2.SpikeMonitor(locals()[pop]) for pop in
---> 63                 record_pops}
     64     # DEBUG ###################
     65     #monitors = utils.record(['pe_mt'], 'spikes', None, None, pe_mt, None, None)

/home/daniel/Science/dopa_net/brian/ardid/tests.py in <dictcomp>((pop,))
     60     # DEBUG ###################
     61     record_pops = ['pe_mt']
---> 62     return {'mon_' + pop: brian2.SpikeMonitor(locals()[pop]) for pop in
     63                 record_pops}
     64     # DEBUG ###################

KeyError: 'pe_mt'

Answer 1

一：忘记eval，因为如果传递给它的字符串是表达式或函数调用，而不是标识符，它可能会导致意想不到的事情发生。如果你真的需要通过名称获取局部变量，你可以使用locals()[name].

干净地完成它

文档：locals

二：所有推导式和生成器表达式（python2.x中的列表推导式除外）have their own namespace, so locals() inside the comprehension will refer to that one - the one that doesn't have your variable. Same goes for eval that captures your local variables by default:

If the locals dictionary is omitted it defaults to the globals dictionary. If both dictionaries are omitted, the expression is executed in the environment where eval() is called.

您可以通过提前获取它们来解决这个问题：

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    groups = locals()    
    return {'mon_' + pop: brian2.SpikeMonitor(eval(pop, globals(), groups)) for pop in record_pops}
    # or better
    return {'mon_' + pop: brian2.SpikeMonitor(groups[pop]) for pop in record_pops}

或者更传统，没有locals：

def test_record():
    groups = {
        "pe_mt": brian2.PoissonGroup(1, 100 * brian2.Hz),
    }
    return {'mon_' + key: brian2.SpikeMonitor(value) for key, value in groups.iteritems()}

Answer 2

不推荐的解决方法：

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    my_loc = locals()
    return {'mon_' + pop: brian2.SpikeMonitor(eval(my_loc[pop])) for pop in
            record_pops}

或者使用普通循环来构建你的字典：

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    d = {}
    for pop in record_pops:
        d['mon_' + pop] = brian2.SpikeMonitor(locals()[pop]))
    return d

或者简单地使用字典来保存对象：

def test_record():
    d = {"pe_mt":brian2.PoissonGroup(1, 100 * brian2.Hz)}
    record_pops = ['pe_mt']
    return {'mon_' + pop: brian2.SpikeMonitor(d[pop]) for pop in record_pops}

字典理解产生看似没有根据的 NameError

Dict comprehension produces seemingly unwarranted NameError

python

scope

list-comprehension

python-internals

dictionary-comprehension