此方法是两个表构建某一相同字段，然后全连接，在做匹配结果筛选，此方法针对数据量不大的时候，逻辑比较简单，但是内存消耗较大
import pandas as pd import numpy as np import re
#关键词数据
df_keyword = pd.DataFrame({
"keyid" : np.arange(5),
"keyword" : ["numpy", "pandas", "matplotlib", "sklearn", "tensorflow"]
})
df_keyword
df_sentence = pd.DataFrame({
"senid" : np.arange(10,17),
"sentence" : [
"怎样用pandas实现merge？",
"Python之Numpy详细教程",
"怎么使用Pandas批量拆分与合并Excel文件？",
"怎样使用pandas的map和apply函数？",
"深度学习之tensorflow简介",
"tensorflow和numpy的关系",
"基于sklearn的一些机器学习的代码"
]
})
df_sentence
df_keyword['match'] = 1 df_sentence['match'] = 1
df_merge = pd.merge(df_keyword, df_sentence) df_merge
def match_func(row):
return re.search(row["keyword"], row["sentence"], re.IGNORECASE) is not None
df_merge[df_merge.apply(match_func, axis = 1)]
匹配结果如下
此方法对编程能力有要求，在大数据集上计算量较方法一小很多
key_word_dict = {
row.keyword : row.keyid
for row in df_keyword.itertuples()
}
key_word_dict
{'numpy': 0, 'pandas': 1, 'matplotlib': 2, 'sklearn': 3, 'tensorflow': 4}
def merge_func(row):
#新增一列，表示可以匹配的keyid
row["keyids"] = [
keyid
for key_word, keyid in key_word_dict.items()
if re.search(key_word, row["sentence"], re.IGNORECASE)
]
return row
df_merge = df_sentence.apply(merge_func, axis = 1)
df_merge
df_result = pd.merge(
left = df_merge.explode("keyids"),
right = df_keyword,
left_on = "keyids",
right_on = "keyid")
df_result
