大数据优化

Question

我在here中发布了一个代码以供审阅。然而，根据现在它没有收到正确的回应，我认为这是由于代码的长度。在这里我将把它切中要害。假设我们有以下列表：

t0=[('Albania','Angola','Germany','UK'),('UK','France','Italy'),('Austria','Bahamas','Brazil','Chile'),('Germany','UK'),('US')]
t1=[('Angola', 'UK'), ('Germany', 'UK'), ('UK', 'France'), ('UK', 'Italy'), ('France', 'Italy'), ('Austria', 'Bahamas')]
t2=[('Angola:UK'), ('Germany:UK'), ('UK:France'), ('UK:Italy'), ('France:Italy'), ('Austria:Bahamas')]

目标是t1中的每一对，我们经过t0，如果找到这一对，我们就用相应的t3元素替换它，我们可以使用以下方法：

result = []
for v1, v2 in zip(t1, t2):
    out = []
    for i in t0:
        common = set(v1).intersection(i)
        if set(v1) == common:
            out.append(tuple(list(set(i) - common) + [v2]))
        else:
            out.append(tuple(i))
    result.append(out)

pprint(result, width=100)  

它给出：

[[('Albania', 'Germany', 'Angola:UK'),
  ('UK', 'France', 'Italy'),
  ('Austria', 'Bahamas', 'Brazil', 'Chile'),
  ('Germany', 'UK'),
  ('U', 'S')],
 [('Albania', 'Angola', 'Germany:UK'),
  ('UK', 'France', 'Italy'),
  ('Austria', 'Bahamas', 'Brazil', 'Chile'),
  ('Germany:UK',),
  ('U', 'S')],
 [('Albania', 'Angola', 'Germany', 'UK'),
  ('Italy', 'UK:France'),
  ('Austria', 'Bahamas', 'Brazil', 'Chile'),
  ('Germany', 'UK'),
  ('U', 'S')],
 [('Albania', 'Angola', 'Germany', 'UK'),
  ('France', 'UK:Italy'),
  ('Austria', 'Bahamas', 'Brazil', 'Chile'),
  ('Germany', 'UK'),
  ('U', 'S')],
 [('Albania', 'Angola', 'Germany', 'UK'),
  ('UK', 'France:Italy'),
  ('Austria', 'Bahamas', 'Brazil', 'Chile'),
  ('Germany', 'UK'),
  ('U', 'S')],
 [('Albania', 'Angola', 'Germany', 'UK'),
  ('UK', 'France', 'Italy'),
  ('Brazil', 'Chile', 'Austria:Bahamas'),
  ('Germany', 'UK'),
  ('U', 'S')]]

此列表的长度为6，表示t1和t2中有6个元素，每个子列表有5个元素，对应于t0中的元素数。从目前的情况来看，代码很快，但在我的例子中，我有t0，它的长度约为48000，t1的长度约为30000。运行时间几乎是永远的我想知道如何用更快的方法执行相同的操作？你知道吗

Answer 1

你可以用一个双重列表来理解。该代码的运行速度约为3.47倍（13.3µs vs 46.2µs）。你知道吗

t0=[('Albania','Angola','Germany','UK'),('UK','France','Italy'),('Austria','Bahamas','Brazil','Chile'),('Germany','UK'),('US')]
t1=[('Angola', 'UK'), ('Germany', 'UK'), ('UK', 'France'), ('UK', 'Italy'), ('France', 'Italy'), ('Austria', 'Bahamas')]
t2=[('Angola:UK'), ('Germany:UK'), ('UK:France'), ('UK:Italy'), ('France:Italy'), ('Austria:Bahamas')]

# We transform the lists of tuple to lists of sets for easier and faster computations
# We transform the lists of tuple to lists of sets for easier and faster computations
t0 = [set(x) for x in t0]
t1 = [set(x) for x in t1]

# We define a function that removes list of elements and adds an element
# from a set 
def add_remove(set_, to_remove, to_add):
    result_temp = set_.copy()
    for element in to_remove:
        result_temp.remove(element)
    result_temp.add(to_add)
    return result_temp

# We do the computation using a double list comprehension
result = [[add_remove(y, x, z) if x.issubset(y) else y for y in t0] 
          for x, z in zip(t1, t2)]