如何有效地检查元素是否在 python 的列表列表中

Question

我有一个列表列表如下。

mylist = [[5274919, ["report", "porcelain", "firing", "technic"]], [5274920, ["implantology", "dentistry"]], [52749, ["method", "recognition", "long", "standing", "root", "perforation", "molar"]], [5274923, ["exogenic", "endogenic", "cause", "tooth", "jaw", "anomaly", "method", "method", "standing"]]]

我还有一个概念列表，如下所示。

myconcepts = ["method", "standing"]

我想看看myconcepts中的每个概念在mylist记录中出现了多少次。 IE;

"method" = 2 times in records (i.e. in `52749` and `5274923`)
"standing" = 2 times in records

我当前的代码如下。

mycounting = 0
for concept in myconcepts:
  for item in mylist:
     if concept in item[1]:
       mycounting = mycounting + 1
print(mycounting)

但是，我当前的mylist非常大，大约有 500 万条记录。 myconcepts列表有大约 10000 个概念。

在我当前的代码中，一个概念需要将近 1 分钟才能获得count ，这非常慢。

我想知道在 python 中执行此操作的最有效方法？

出于测试目的，我将一小部分数据集附加到： https://drive.google.com/file/d/1z6FsBtLyDZClod9hK8nK4syivZToa7ps/view?usp=sharing

如果需要，我很乐意提供更多详细信息。

Answer 1

您可以展平输入，然后使用collections.Counter ：

import collections
myconcepts = ["method", "standing"]
mylist = [[5274919, ["report", "porcelain", "firing", "technic"]], [5274920, ["implantology", "dentistry"]], [5274921, ["method", "recognition", "long", "standing", "root", "perforation", "molar"]], [5274923, ["exogenic", "endogenic", "cause", "tooth", "jaw", "anomaly", "method", "standing"]]]
def flatten(d):
  for i in d:
    yield from [i] if not isinstance(i, list) else flatten(i)

r = collections.Counter(flatten(mylist))
result = {i:r.get(i, 0) for i in myconcepts}

Output：

{'method': 2, 'standing': 2}

---

编辑：记录查找：

result = {i:sum(i in b for _, b in mylist) for i in myconcepts}

Output：

{'method': 2, 'standing': 2}

Answer 2

从https://www.geeksforgeeks.org/python-count-the-sublists- contains-given-element-in-a-list/调整方法 3

from itertools import chain 
from collections import Counter 

mylist = [[5274919, ["report", "porcelain", "firing", "technic"]], [5274920, ["implantology", "dentistry"]], [52749, ["method", "recognition", "long", "standing", "root", "perforation", "molar"]], [5274923, ["exogenic", "endogenic", "cause", "tooth", "jaw", "anomaly", "method", "method", "standing"]]]

myconcepts = ["method", "standing"]

def countList(lst, x):
" Counts number of times item x appears in sublists "
    return Counter(chain.from_iterable(set(i[1]) for i in lst))[x] 

# Use dictionary comprehension to apply countList to concept list
result = {x:countList(mylist, x) for x in myconcepts}
print(result) # {'method':2, 'standing':2}

*修改当前方法（只计算一次）*

def count_occurences(lst):
    " Number of counts of each item in all sublists "
    return Counter(chain.from_iterable(set(i[1]) for i in lst))

cnts = count_occurences(mylist)
result = {x:cnts[x] for x in myconcepts}
print(result) # {'method':2, 'standing':2}

性能（比较使用 Jupyter Notebook 发布的方法）

结果显示此方法与 Barmar 发布的方法接近（即 36 与 42 us）

对当前方法的改进将时间缩短了大约一半（即从 36 微秒减少到 19 微秒）。 对于更多的概念（即问题有 > 1000 个概念），这种改进应该更加显着。

但是，原始方法更快，为 2.55 us/loop。

方法 当前方法

%timeit { x:countList(mylist, x) for x in myconcepts}
#10000 loops, best of 3: 36.6 µs per loop

Revised current method:

%%timeit
cnts = count_occurences(mylist)
result = {x:cnts[x] for x in myconcepts}
10000 loops, best of 3: 19.4 µs per loop

方法2（来自Barmar post）

%%timeit
r = collections.Counter(flatten(mylist))
{i:r.get(i, 0) for i in myconcepts}
# 10000 loops, best of 3: 42.7 µs per loop

方法3（原始方法）

%%timeit

result = {}
for concept in myconcepts:
  mycounting = 0
  for item in mylist:
     if concept in item[1]:
       mycounting = mycounting + 1
  result[concept] = mycounting
  # 100000 loops, best of 3: 2.55 µs per loop

Answer 3

将概念列表更改为集合，以便搜索将是 O(1)。 然后，您可以使用交集来计算每组中的匹配数。

import set
mylist = [
    [5274919, {"report", "porcelain", "firing", "technic"}], 
    [5274920, {"implantology", "dentistry"}], 
    [52749, {"method", "recognition", "long", "standing", "root", "perforation", "molar"}], 
    [5274923, {"exogenic", "endogenic", "cause", "tooth", "jaw", "anomaly", "method", "method", "standing"}]
]
myconcepts = {"method", "standing"}
mycounting = 0
for item in mylist:
    mycounting += len(set.intersection(myconcepts, item[1]))
print(mycounting)

如果要分别获取每个概念的计数，则需要遍历myconcept ，然后使用in运算符。 您可以将结果放入字典中。

mycounting = {concept: sum(1 for l in mylist if concept in l[1]) for concept in myconcepts}
print(mycounting) // {'standing': 2, 'method': 2}

这仍然比使用列表更有效，因为concept in l[1]是 O(1)。

Answer 4

据我所知，在 python 中编程非常慢，因为编译器非常懒惰。 所以，我认为没有简单的方法来解决这个问题，或者通过改变计算机语言。