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获取出现在3个或更多列表中的元素

comparison list python

Mechanician · 技术社区 · 8 年前

比如说,我总共有5个列表

# Sample data

a1 = [1,2,3,4,5,6,7]

a2= [1,21,35,45,58]
a3= [1,2,15,27,36]
a4=[2,3,1,45,85,51,105,147,201]
a5=[3,458,665]

我需要找到a1的元素,这些元素在a2、a3、a4、a5中的存在次数超过3次,包括a1中的元素

或

我需要所有列表(a1-a5)中频率大于或等于3的元素,以及它们的频率。

从上面的示例中,预期输出为:

2,频率为3

3,频率为3

对于我的实际问题,列表的数量和长度都是如此巨大,有人能给我一个简单快速的方法吗?

普利提维

3 回复 | 直到 8 年前

jme 8 年前

帕特里克在评论中写道, chain 和 Counter 你的朋友在这里吗

import itertools
import collections

targets = [1,2,3,4,5,6,7]

lists = [
    [1,21,35,45,58],
    [1,2,15,27,36],
    [2,3,1,45,85,51,105,147,201],
    [3,458,665]
    ]

chained = itertools.chain(*lists)
counter = collections.Counter(chained)
result = [(t, counter[t]) for t in targets if counter[t] >= 2]

如此

>>> results
[(1, 3), (2, 2), (3, 2)]

你说你有很多列表,每个列表都很长。尝试这个解决方案,看看需要多长时间。如果需要加快,那是另一个问题。可能是这样 collections.Counter 对于您的应用程序来说太慢。

Whud 8 年前

a1= [1,2,3,4,5,6,7]
a2= [1,21,35,45,58]
a3= [1,2,15,27,36]
a4= [2,3,1,45,85,51,105,147,201]
a5= [3,458,665]

b = a1+a2+a3+a4+a5                              #make b all lists together

for x in set(b):                                #iterate though b's set
    print(x, 'with a frequency of', b.count(x)) #print the count

将为您提供:

1 with a frequency of 4
2 with a frequency of 3
3 with a frequency of 3
4 with a frequency of 1
5 with a frequency of 1
6 with a frequency of 1
7 with a frequency of 1
35 with a frequency of 1
36 with a frequency of 1
...

编辑:

使用:

for x in range(9000):
    a1.append(random.randint(1,10000))
    a2.append(random.randint(1,10000))
    a3.append(random.randint(1,10000))
    a4.append(random.randint(1,10000))

time 我检查了程序花费了多长时间(不打印,而是保存信息),程序花费了4.9395秒。我希望这足够快。

VersBersch 8 年前

使用熊猫的解决方案相当快

import pandas as pd

a1=[1,2,3,4,5,6,7]
a2=[1,21,35,45,58]
a3=[1,2,15,27,36]
a4=[2,3,1,45,85,51,105,147,201]
a5=[3,458,665]

# convert each list to a DataFrame with an indicator column
A = [a1, a2, a3, a4, a5]
D = [ pd.DataFrame({'A': a, 'ind{0}'.format(i):[1]*len(a)}) for i,a in enumerate(A)]

# left join each dataframe onto a1
# if you know the integers are distinct then you don't need drop_duplicates
df = pd.merge(D[0], D[1].drop_duplicates(['A']), how='left', on='A')
for d in D[2:]:
    df = pd.merge(df, d.drop_duplicates(['A']), how='left', on='A')

# sum accross the indicators
df['freq'] = df[['ind{0}'.format(i) for i,d in enumerate(D)]].sum(axis=1)

# drop frequencies less than 3
print df[['A','freq']].loc[df['freq'] >= 3]

在我的机器上,使用以下较大输入的测试运行时间不到0.2秒

import numpy.random as npr
a1 = xrange(10000)
a2 = npr.randint(10000, size=100000) 
a3 = npr.randint(10000, size=100000) 
a4 = npr.randint(10000, size=100000) 
a5 = npr.randint(10000, size=100000)