在一只新的熊猫数据中逐行遍历一列列表，并在X上翻转匹配

Question

我正在尝试迭代在熊猫dataframe列中找到的列表，并在新的dataframe中返回包含在其他行中的三次以上匹配的结果。

，以下是数据的外观：

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期望输出：

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(输出是因为这些特定关键字在列表中至少在另外三行中找到)。

最小可重现性示例：

import pandas as pd

# initialize data of lists.
data = {'url': ["www.bbc.co.uk", "www.cabinzero.com", "www.cntraveller.com", "www.forbes.com", "www.gov.scot", "www.gov.uk", "www.ons.gov.uk"],
        'keyword': ["['amber travel list', 'travel amber list', 'amber list countries uk travel', 'travel amber list countries', 'amber list countries travel']", "['amber list countries uk travel', 'travel amber list countries', 'amber travel list', 'travel amber list', 'amber list countries travel']", "['travel amber list', 'amber list countries uk travel', 'amber travel list', 'amber list countries travel', 'travel amber list countries']", "['amber travel list', 'travel amber list countries', 'travel amber list', 'amber list countries travel', 'amber list countries uk travel']", "['amber list countries travel', 'travel amber list countries', 'amber list countries uk travel', 'travel amber list', 'amber travel list']", "['amber list countries travel', 'amber list countries uk travel', 'amber travel list']", "['amber list countries uk travel', 'amber travel list', 'travel amber list countries', 'amber list countries travel']"]}

# Create DataFrame
df = pd.DataFrame(data)

# Print the output.
print(df)

--我已经尝试过了--我尝试过将列表列转储到一个列表中，然后迭代来计算出现的次数，但是不能让它工作，并且不确定这是否是最好的方法。

Answer 1

如果在同一个列表中，每个关键字都是唯一的，那么您可以：

from itertools import chain

listed_keywords = df.keyword.apply(lambda x: eval(x)).values # returns array of list
all_keywords = list(chain.from_iterable(listed_keywords)) # Concat all the lists into 1 global list of keywords

unique_keyword, nunique_keyword = np.unique(all_keywords, return_counts = True)# Return unique keywords and their respective frequency among all the keywords

df_keywords = pd.DataFrame(dict(keyword = unique_keyword, frequency = nunique_keyword)) # Create a DataFrame so you can easily filter according to keyword frequency.

希望这回答了你的问题！