在 Python 中用另一个元组过滤一个元组

Question

我有一个使用zip 函数创建的元组列表。 zip 汇集了四个列表：narrative、subject、activity 和 filer，每个列表都只是 0 和 1 的列表。假设这四个列表如下所示：

narrative = [0, 0, 0, 0]
subject = [1, 1, 0, 1]
activity = [0, 0, 0, 1]
filer = [0, 1, 1, 0]

现在，我将 zip 将它们放在一起以获取布尔值列表，指示它们中是否有任何 True。

ny_nexus = [True if sum(x) > 0 else False for x in zip(narrative, subject, activity, filer)]

我现在遇到的问题是获取第二个元组列表，如果它在迭代期间为 1，则返回变量的名称。我想它看起来像这样：

variables = ("narrative", "subject", "activity", "filer")
reason = [", ".join([some code to filter a tuple]) for x in zip(narrative, subject, activity, filer)]

我只是不知道该怎么做。我想要的输出如下所示：

reason
# ["subject", "subject, filer", "filer", "subject, activity"]

我对 Python 有点陌生，所以如果解决方案很简单，我深表歉意。

Answer 1

将元组存储在字典中以获得更简洁的解决方案：

tups = {'narrative': narrative,
        'subject': subject,
        'activity': activity,
        'filer': filer}

解决办法：

reason = [', '.join(k for k, b in zip(tups, x) if b) for x in zip(*tups.values())]

也可以用itertools.compress写：

from itertools import compress
reason = [', '.join(compress(tups, x)) for x in zip(*tups.values())]

---

上述解决方案不保留元组的顺序，例如他们可以返回类似的东西

['subject', 'filer, subject', 'filer', 'activity, subject']

如果您需要保留订单，请使用collections.OrderedDict，如下所示：

from collections import OrderedDict

tups = OrderedDict([
    ('narrative', narrative),
    ('subject', subject),
    ('activity', activity),
    ('filer', filer)
])

# The result is ['subject', 'subject, filer', 'filer', 'subject, activity']

编辑：不涉及字典的解决方案：

from itertools import compress
reason = [', '.join(compress(variables, x))
          for x in zip(narrative, subject, activity, filer)]

如果zip(...) 调用不再适合一行，请考虑使用字典。

Answer 2

使用zip(narrative, subject, activity, filer) 基本上会转置矩阵（您的等长列表组成矩阵）。然后您通过这些枚举找到标志为真的位置n 并索引适当的变量。

narrative = [0, 0, 0, 0]
subject = [1, 1, 0, 1]
activity = [0, 0, 0, 1]
filer = [0, 1, 1, 0]
variables = ("narrative", "subject", "activity", "filer")
# ========================================================

new_list = [[variables[n] for n, flag in enumerate(indicators) if flag] 
            for indicators in zip(narrative, subject, activity, filer)]
>>> new_list
[['subject'], ['subject', 'filer'], ['filer'], ['subject', 'activity']]

要查看转置：

>>> [i for i in zip(narrative, subject, activity, filer)]

Answer 3

您可以只使用理解语法的过滤方面来获取 仅当相应标志为 True 时，您的可用英文名称：

variables = ("narrative", "subject", "activity", "filer")
[tuple (name for flag, name in zip(x, variables) if x)  for x in zip(narrative, subject, activity, filer)]

也就是说，您的方法有些可疑 - 您可能会（远远）更好地使用面向对象的方法，而不是尝试手动协调每个主题的独立变量序列。

Answer 4

    narrative = [0, 0, 0, 0]
    subject = [1, 1, 0, 1]
    activity = [0, 0, 0, 1]
    filer = [0, 1, 1, 0]
    variables = ("narrative", "subject", "activity", "filer")
    ny_nexus = [True if sum(x) > 0 else False for x in zip(narrative, subject, activity, filer)]
    output = []
    [[output.append(variables[j]) if t==1 else None for j,t in enumerate(x)] for x in zip(narrative, subject, activity, filer)]
    print ny_nexus
    print output

当然，您可以在不使用列表推导的情况下执行以下操作：

    narrative = [0, 0, 0, 0]
    subject = [1, 1, 0, 1]
    activity = [0, 0, 0, 1]
    filer = [0, 1, 1, 0]
    variables = ("narrative", "subject", "activity", "filer")
    ny_nexus = [True if sum(x) > 0 else False for x in zip(narrative, subject, activity, filer)]
    output = []
    for x in zip(narrative, subject, activity, filer):
        for j,t in enumerate(x):
            output.append(variables[j])
    print ny_nexus
    print output