从dataframe - Pandas的列中筛选出关键字(不区分大小写)

Question

想要根据现有列中的单词创建一个新列，新列应该是Lobby、UPS、Electrical或“空白空间”

Name                                SubUnitName
Lobby Area                          Lobby
Sensor - Bank lobby                 Lobby
Temperature - UPS Room              UPS
Sensor - Electric Room              Electric
Sensor - electrical Room            Electric
Temperature - electric Room         Electric
Sensor

如上所示，搜索应该是不区分大小写的，如果找到“电气”或“电气”，则结果应该是“电气”

Answer 1

建立要在"Name“列中查找的单词列表，然后应用函数"find_match”来创建新的"SubUnitName“列。

search_list = ["Lobby", "UPS", "Electric"]


def find_match(name_str: str) -> str:
    for item in search_list:
        item_lc = item.lower()
        if item_lc in name_str.lower():
            return item


df.loc[:, "SubUnitName"] = df["Name"].apply(find_match)

用最后一行的空白替换None

df["SubUnitName"].fillna('', inplace=True)

Answer 2

我为你提供了一个解决方案。它检查字符串之间的匹配，如果找到，则将其添加到将成为您的新列的列表中。

import pandas as pd 


d = { "Name" : ["Sensor - Bank lobby ", "Sensor - Bank Lobby ", "Temperature - UPS Room", "Sensor - Electric Room ", "Sensor - electrical Room", "Sensor"]}


df = pd.DataFrame(data=d)

list_sub_units = []

list_matches = ["Lobby", "UPS", "Electric"]

for entry in df["Name"]:
    matched = False

    for match in list_matches:
        if entry.lower().find(match.lower()) > 0:
            list_sub_units.append(match)
            matched = True
        
    if not matched:
        list_sub_units.append("")

df["SubUnitName"] = list_sub_units


print(df)