Question

我有一个二维列表：

arr = [['Mohit', 'shini','Manoj','Mot'],
      ['Mohit', 'shini','Manoj'],
      ['Mohit', 'Vis', 'Nusrath']]

我想在2D列表中找到最频繁的元素。在上面的示例中，最常见的字符串是'Mohit'。

我知道我可以使用蛮力使用两个for循环和一个字典来执行此操作，但是有没有使用numpy或任何其他库的更有效方法？

嵌套列表的长度可以不同

有人还能添加方法的时间吗？查找禁食方法。还有一些警告，可能会导致效率不高。

修改

这些是我系统上不同方法的时间：

#timegb
%%timeit
collections.Counter(chain.from_iterable(arr)).most_common(1)[0][0]
5.91 µs ± 115 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#Kevin Fang and Curious Mind
%%timeit
flat_list = [item for sublist in arr for item in sublist]
collections.Counter(flat_list).most_common(1)[0]
6.42 µs ± 501 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
c = collections.Counter(item for sublist in arr for item in sublist).most_common(1)c[0][0]
6.79 µs ± 449 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#Mayank Porwal
def most_common(lst):
    return max(set(lst), key=lst.count)
%%timeit
ls = list(chain.from_iterable(arr))
most_common(ls)
2.33 µs ± 42.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#U9-Forward
%%timeit
l=[x for i in arr for x in i]
max(l,key=l.count)
2.6 µs ± 68.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Mayank Porwal的方法在我的系统上运行最快。

Answer 1

用itertools.chain.from_iterable整理列表
应用Counter。

演示：

>>> from itertools import chain
>>> from collections import Counter
>>> 
>>> lst = [['Mohit', 'shini','Manoj','Mot'],
...:      ['Mohit', 'shini','Manoj'],
...:      ['Mohit', 'Vis', 'Nusrath']]
...:      
>>> Counter(chain.from_iterable(lst)).most_common(1)[0][0]
'Mohit'

详细信息：

>>> list(chain.from_iterable(lst))
['Mohit',
 'shini',
 'Manoj',
 'Mot',
 'Mohit',
 'shini',
 'Manoj',
 'Mohit',
 'Vis',
 'Nusrath']
>>> Counter(chain.from_iterable(lst))
Counter({'Manoj': 2, 'Mohit': 3, 'Mot': 1, 'Nusrath': 1, 'Vis': 1, 'shini': 2})
>>> Counter(chain.from_iterable(lst)).most_common(1)
[('Mohit', 3)]

一些时间：

>>> lst = lst*100
>>> %timeit Counter(chain.from_iterable(lst)).most_common(1)[0][0] # timgeb
53.7 µs ± 411 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit max([x for i in lst for x in i], key=l.count) # U9-Forward
207 µs ± 389 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit Counter([x for sublist in lst for x in sublist]).most_common(1)[0][0] # Curious_Mind/Kevin Fang #1
75.2 µs ± 2.6 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit Counter(item for sublist in lst for item in sublist).most_common(1)[0][0] # Kevin Fang #2
95.2 µs ± 2.07 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit flat = list(chain.from_iterable(lst)); max(set(flat), key=flat.count) # Mayank Porwal
98.4 µs ± 178 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

（请注意，Kevin Fang的第二个解决方案比第一个解决方案要慢一些，但内存效率更高。）

Answer 2

我建议展平2D数组，然后使用计数器找出最频繁的元素。

flat_list = [item for sublist in arr for item in sublist]
from collections import Counter
Counter(flat_list).most_common(1)[0]
# ('Mohit', 3)
Counter(flat_list).most_common(1)[0][0]
# 'Mohit'

不确定这是否是最快的方法。

编辑：

@timgeb的答案可以使用itertools.chain

来更快地整理列表

@schwobaseggl建议的一种更节省空间的方法：

from collections import Counter
c = Counter(item for sublist in arr for item in sublist).most_common(1)
# [('Mohit', 3)]
c[0][0]
# 'Mohit'

Answer 3

类似这样的东西：

In [920]: from itertools import chain
In [923]: arr = list(chain.from_iterable(arr)) ## flatten into 1-D array
In [922]: def most_common(lst):
     ...:     return max(set(lst), key=lst.count)

In [924]: most_common(arr)
Out[924]: 'Mohit'

时间：

from itertools import chain
import time
start_time = time.time()

arr = [['Mohit', 'shini','Manoj','Mot'],
      ['Mohit', 'shini','Manoj'],
      ['Mohit', 'Vis', 'Nusrath']]


arr = list(chain.from_iterable(arr))
arr = arr*100

def most_common(lst):
    return max(set(lst), key=lst.count)

print(most_common(arr))
print("--- %s seconds ---" % (time.time() - start_time))

mayankp@mayank:~$ python t1.py 
Mohit
--- 0.000154972076416 seconds ---

Answer 4

一种这样做的方法，

import collections
import time
start_time = time.time()
arr = [['Mohit', 'shini','Manoj','Mot'],
      ['Mohit', 'shini','Manoj'],
      ['Mohit', 'Vis', 'Nusrath']]

c = collections.Counter([x for sublist in arr for x in sublist])
print(c.most_common(1) )
print("--- %s seconds ---" % (time.time() - start_time))

花费时间： 0.00016713142395秒

演示：http://tpcg.io/NH3zjm

Answer 5

或者为什么不呢？

l=[x for i in arr for x in i]
max(l,key=l.count)

代码示例：

>>> arr = [['Mohit', 'shini','Manoj','Mot'],
      ['Mohit', 'shini','Manoj'],
      ['Mohit', 'Vis', 'Nusrath']]
>>> l=[x for i in arr for x in i]
>>> max(l,key=l.count)
'Mohit'
>>>

在2D列表中查找最常见的字符串

5 个答案: