测试列表列表以根据约束删除不需要的列表

Question

我有一个列表列表。列表由来自某些区域的人组成，如果列表中有来自某个区域的太多人，我想从列表集中删除该列表。列表的长度为 9

list=[[["Aarat","California"],
["Aaron","California"],
["Abba","California"],
["Abaddon","California"],
["Abner","Nevada"],
["Abram","Nevada"],
["Abraham","Nevada"],
["Absalom","Nevada"],
["Adullam","Utah"]],
......,
[["Abital","California"],
["Abitub","California"],
["Absalom","Nevada"],
["Accad","Nevada"],
["Agar","Utah"],
["Agee","Utah"],
["Aijeleth-Shahar","New Mexico"],
["Ain","New Mexico"],
["Amram","Washington"]]]
Cities=["California","Nevada","Utah","New Mexico","Idaho","Washington"]
denk=[] 
for city in Cities:
    den=[]
    for i in list:
        a=i[0]
        b=i[1]
        c=i[2]
        d=i[3]
        e=i[4]
        f=i[5]
        g=i[6]
        h=i[7]
        k=i[8]
        if a==city:
            ab=1
        if b==city:
            ac=1
        if c==city:
            ad=1
        if d==city:
            ae=1
        if e==city:
            af=1
        if f==city:
            ag=1
        if g==city:
            ah=1
        if h==city:
            ai=1
        if k==city:
            aj=1
        if (ab+ac+ad+ae+af+ag+ah+ai+aj)>3:
            den.append(1)
        if (ab+ac+ad+ae+af+ag+ah+ai+aj)<4:
            den.append(0)
    denk.append(sum(den))

finalList=[]
for i, j in enumerate(denk):
    if j == 0:
        finalList.append(list[i])

我尝试计算来自城市的人数，如果人数大于 3，我尝试添加 1，如果不是 0。我这样做只是为了总结列表的次数超过配额。

Cities=["California","Nevada","Utah","New Mexico","Idaho","Washington"]

[["Aarat","California"],
["Aaron","California"],
["Abba","California"],
["Abaddon","California"],
["Abner","Nevada"],
["Abram","Nevada"],
["Abraham","Nevada"],
["Absalom","Nevada"],
["Adullam","Utah"]]

在测试这个特定列表时，测试看看有多少人来自加利福尼亚，因为有超过 3 人来自加利福尼亚，所以 den=1。下一个城市，内华达州，也会使 den=1，依此类推...... den=[1,1,0,0,0,0] 登克=[2] 所以这个列表被扔掉了

[["Abital","California"],
["Abitub","California"],
["Absalom","Nevada"],
["Accad","Nevada"],
["Agar","Utah"],
["Agee","Utah"],
["Aijeleth-Shahar","New Mexico"],
["Ain","New Mexico"],
["Amram","Washington"]]

在此处执行相同操作会为 Cities 中的每个城市生成 den=0，den=[0,0,0,0,0,0]，denk=[0]，因此该列表将被接受。

finalList 不应包含任何来自一个地方的人员过多的列表。

Answer 1

假设您从以下内容开始：

list=[[["Aarat","California"],
    ["Aaron","California"],
    ["Abba","California"],
    ["Abaddon","California"],
    ["Abner","Nevada"],
    ["Abram","Nevada"],
    ["Abraham","Nevada"],
    ["Absalom","Nevada"],
    ["Adullam","Utah"]],[["Abital","California"],
    ["Abitub","California"],
    ["Absalom","Nevada"],
    ["Accad","Nevada"],
    ["Agar","Utah"],
    ["Agee","Utah"],
    ["Aijeleth-Shahar","New Mexico"],
    ["Ain","New Mexico"],
    ["Amram","Washington"]]]

要查找每个二级列表中的分布，您可以使用列表理解和collections.Counter：

import collections

>>> [collections.Counter(e[1] for e in l) for l in list]
[Counter({'California': 4, 'Nevada': 4, 'Utah': 1}),
 Counter({'California': 2,
          'Nevada': 2,
          'New Mexico': 2,
          'Utah': 2,
          'Washington': 1})]

要在每个二级列表中查找最常见的计数，您可以使用

>>> [collections.Counter(e[1] for e in l).most_common(1)[0][1] for l in list]
[4, 2]

因此，要仅保留最常见计数最多为 3 的二级列表，您可以使用

>>> [l for l in list if collections.Counter(e[1] for e in l).most_common(1)[0][1] <= 3]
[[['Abital', 'California'],
  ['Abitub', 'California'],
  ['Absalom', 'Nevada'],
  ['Accad', 'Nevada'],
  ['Agar', 'Utah'],
  ['Agee', 'Utah'],
  ['Aijeleth-Shahar', 'New Mexico'],
  ['Ain', 'New Mexico'],
  ['Amram', 'Washington']]]