在 Pandas 数据框中查找多个字典键 & return 多个匹配值

Question

第一次发帖，格式不对请见谅

这是我的问题：

我创建了一个 Pandas 数据框，其中包含多行文本：

d = {'keywords' :['cheap shoes', 'luxury shoes', 'cheap hiking shoes']}
keywords = pd.DataFrame(d,columns=['keywords'])
In [7]: keywords
Out[7]:
        keywords
0  cheap shoes
1  luxury shoes
2  cheap hiking shoes

现在我有一个包含以下键/值的字典：

labels = {'cheap' : 'budget', 'luxury' : 'expensive', 'hiking' : 'sport'}

我想做的是找出数据框中是否存在字典中的键，如果存在，return 适当的值

我能够使用以下方法到达那里：

for k,v in labels.items():
   keywords['Labels'] = np.where(keywords['keywords'].str.contains(k),v,'No Match')

但是，输出缺少前两个键，只捕获最后一个 "hiking" 键

    keywords            Labels
0   cheap shoes         No Match
1   luxury shoes        No Match
2   cheap hiking shoes  sport

此外，我还想知道是否有办法在字典中捕获多个由 | 分隔的值, 所以理想的输出应该是这样的

    keywords            Labels
0   cheap shoes         budget
1   luxury shoes        expensive
2   cheap hiking shoes  budget | sport

非常感谢任何帮助或指导。

干杯

Answer 1

当然有可能。这是一种方法。

d = {'keywords': ['cheap shoes', 'luxury shoes', 'cheap hiking shoes', 'nothing']}

keywords = pd.DataFrame(d,columns=['keywords'])

labels = {'cheap': 'budget', 'luxury': 'expensive', 'hiking': 'sport'}

df = pd.DataFrame(d)

def matcher(k):
    x = (i for i in labels if i in k)
    return ' | '.join(map(labels.get, x))

df['values'] = df['keywords'].map(matcher)

#              keywords          values
# 0         cheap shoes          budget
# 1        luxury shoes       expensive
# 2  cheap hiking shoes  budget | sport
# 3             nothing

Answer 2

您可以使用 "|".join(labels.keys()) 获取 re.findall() 使用的模式。

import pandas as pd
import re

d = {'keywords' :['cheap shoes', 'luxury shoes', 'cheap hiking shoes']}
keywords = pd.DataFrame(d,columns=['keywords'])
labels = {'cheap' : 'budget', 'luxury' : 'expensive', 'hiking' : 'sport'}
pattern = "|".join(labels.keys())

def f(s):
    return "|".join(labels[word] for word in re.findall(pattern, s))

keywords.keywords.map(f)

Answer 3

坚持你的方法，你可以做例如

arr = np.array([np.where(keywords['keywords'].str.contains(k), v, 'No Match') for k, v in labels.items()]).T
keywords["Labels"] = ["|".join(set(item[ind if ind.sum() == ind.shape[0] else ~ind])) for item, ind in zip(arr, (arr == "No Match"))]

Out[97]: 
             keywords        Labels
0         cheap shoes        budget
1        luxury shoes     expensive
2  cheap hiking shoes  sport|budget

Answer 4

您可以split the strings into separate columns, then stack into a multi index, so that you can map, the labels dictionary to the values. Then groupby the initial index, and concatenate属于每个索引的字符串

keywords['Labels'] = keywords.keywords.str.split(expand=True).stack()\
                     .map(labels).groupby(level=0)\
                     .apply(lambda x: x.str.cat(sep=' | '))



            keywords          Labels
0         cheap shoes          budget
1        luxury shoes       expensive
2  cheap hiking shoes  budget | sport

Answer 5

我喜欢先使用 replace 然后找到值的想法。

keywords.assign(
    values=
    keywords.keywords.replace(labels, regex=True)
            .str.findall(f'({"|".join(labels.values())})')
            .str.join(' | ')
)

             keywords          values
0         cheap shoes          budget
1        luxury shoes       expensive
2  cheap hiking shoes  budget | sport

在 Pandas 数据框中查找多个字典键 & return 多个匹配值

Looking up multiple dictionary keys in a Pandas Dataframe & return multiple values for matches

python

dictionary

string-matching

python-3.x

pandas