如何基于多个已排序项目列表来创建已排序项目列表？

Question

我正在尝试从较小的列表中获取经过排序的主列表。可以使用python或R。

在R中，我有

l1<-c("a","c","d")
l2<-c("a","b","e")
l3<-c("a","c","e")
l4<-c("a","b","c","e")
l5<-c("b","c","d")

m<-unique(c(l1,l2,l3,l4,l5))

我期望的输出是a,b,c,d,e。

在python中

l1=["a","c","d"]
l2=["a","b","e"]
l3=["a","c","e"]
l4=["a","b","c","e"]
l5=["b","c","d"]

预期结果为["a","b","c","d","e"] 我首先进行设置并遍历每个列表并检查索引，但是开始变得复杂。

感谢您的帮助。

编辑： 我想我对按字母顺序排序的列表感到困惑。这些列表项可能是随机的。实际订单可能不是a,b,c,d,e。

l1=["e","a"]
l2=["e","b","d"]
l3=["b","d","a"]

在这种情况下，预期顺序为["e","b","d","a"]而不是["a","b","d","e"]

为更清楚起见，请考虑很少有人试图从东到西命名美国各州。

person 1 says, Florida, Louisiana, Nevada,California. 
person 2 says  Alabama, Mississippi, Louisiana, new Mexico, Nevada 
person 3 says Florida, Alabama, Texas, New Mexico, California
person 4 says Alabama, Mississippi, Texas, Nevada
person 5 says Mississippi Louisiana, Nevada

我正在尝试从上述信息中获得正确的顺序。

因此，在这里，我们将从Florida, Louisiana, Nevada, California开始。现在，加第二，就是(Alabama, Florida),Louisiana,New Mexico, Nevada, California.加第三（打破阿拉巴马州/佛罗里达的平局），Florida, Alabama, Louisiana, Texas, New Mexico, Nevada, California并加第四使Florida, Alabama, (Mississippi/Louisiana), Texas, New Mexico, Nevada, California。密西西比州和路易斯安那州增加了第5个突破领带。

Answer 1

Aaa和实际答案：https://www.python.org/doc/essays/graphs/

打猎好！ ：D

这应该针对您的原始问题：

l1=["a","c","d"]
l2=["a","b","e"]
l3=["a","c","e"]
l4=["a","b","c","e"]
l5=["b","c","d"]

s = set()
s.update(l1, l2, l3, l4, l5)
l = sorted(s)
l
#['a', 'b', 'c', 'd', 'e']

对于您已编辑的问题，让我们考虑第二个示例的细微变化：

l1=["e","a"]
l2=["e","b","d"]
l3=["b","c","a"]

（在l3上斜视）。在这种情况下，列表集未充分确定，因为d和c之间没有唯一的顺序。没有确定联系的规则，就不可能有算法。

Answer 2

对于Python：

# Create list of lists
lsts = [l1, l2, l3, l4, l5]
s = set()

# Add lists to set
for lst in lsts:
  s.update(lst)

# Sort set
sorted(s)

编辑：OP更新后：

def sort_lists(lsts):
  list_of_hashes = []
  for lst in lsts:
    list_of_hashes.append({k: v for v, k in enumerate(lst)})

  result_hash = dict()
  for hash_item in list_of_hashes:
    for key, value in hash_item.items():
      if result_hash.get(key):
        result_hash[key] += value
      else:
        result_hash[key] = value
  print(result_hash)
  sorted_results = sorted(result_hash.items(), key=lambda kv: kv[1])
  print(sorted_results)
  return [tup[0] for tup in sorted_results]

# Test Case 1
l1=["e","a"]
l2=["e","b","d"]
l3=["b","d","a"]

print(sort_lists([l1,l2,l3]))
>> ['e', 'd', 'b', 'a']

# Test Case 2
s1 = ['Florida', 'Louisiana', 'Nevada', 'California']
s2 = ['Alabama', 'Mississippi', 'Louisiana', 'New Mexico', 'Nevada']
s3 = ['Florida', 'Alabama', 'Texas', 'New Mexico', 'California']
s4 = ['Alabama', 'Mississippi', 'Texas', 'Nevada']
s5 = ['Mississippi', 'Louisiana', 'Nevada']
print(sort_lists([s1,s2,s3,s4,s5]))
>> ['Florida', 'Alabama', 'Mississippi', 'Louisiana', 'Texas', 'New Mexico', 'California', 'Nevada']

Answer 3

这是R中的一种方法，该方法使用tidygraph将向量转换为有向无环图，然后使用node_topo_order得出隐含的节点顺序。使用从东到西的示例状态：

l1 <- c("Florida", "Louisiana", "Nevada", "California")
l2 <- c("Alabama", "Mississippi", "Louisiana", "New Mexico", "Nevada" )
l3 <- c("Florida", "Alabama", "Texas", "New Mexico", "California")
l4 <- c("Alabama", "Mississippi", "Texas", "Nevada")
l5 <- c("Mississippi", "Louisiana", "Nevada")

library(tidyverse)
library(tidygraph)
ew_graph <- list(l1, l2, l3, l4, l5) %>%
  map_dfr(~tibble(east = ., west = lead(.))) %>% # turn vectors into edge table
  filter(!is.na(west)) %>%
  as_tbl_graph()

ew_graph %>%  # Now we can order nodes and extract their names as output
  arrange(node_topo_order()) %>%
  pull(name)
#> [1] "Florida"     "Alabama"     "Mississippi" "Louisiana"   "Texas"      
#> [6] "New Mexico"  "Nevada"      "California"

请注意，可以有多个正确的订单，这只会返回其中之一。如果需要，我们还可以绘制图形以更清楚地查看它们之间的关系，这表明在此数据中，路易斯安那州和德克萨斯州之间存在联系（您无法相互追踪），而在构建例。碰巧我们按“真实”顺序拥有它们。如果您需要定义一种单独的打破关系的方式，则这种方法将需要一些技巧。

library(ggraph)
ggraph(ew_graph) +
  geom_node_label(aes(label = name)) +
  geom_edge_link(
    mapping = aes(start_cap = label_rect(node1.name),
                  end_cap = label_rect(node2.name)),
    arrow = arrow(length = unit(4, 'mm'))
  )

^{由reprex package（v0.3.0）于2019-05-28创建}

Answer 4

我的解决方案的复杂度为 O（n）。其他解决方案可以具有O（n log n）：

Python ：（与R类似）

l1=["a","c","d"]
l2=["a","b","e"]
l3=["a","c","e"]
l4=["a","b","c","e"]
l5=["b","c","d"]

lsts = [l1, l2, l3, l4, l5]

solve = []
for p in range(130):
    solve.append(0)

for lst in lsts:
    for p in lst:
        solve[ord(p)] += 1

for idx, value in enumerate(solve):
    if value != 0:
        print chr(idx)

此解决方案基于表ascii中的值。

对于您的更新：

l1=["z","c","d"]
l2=["a","b","e"]
l3=["a","c","e"]
l4=["a","b","c","e"]
l5=["b","c","d"]

mySet = set()

mySet.update(l1, l2, l3, l4, l5)

result = sorted(mySet)
print(result)