Python 如何拆分列中的字符串值

发布于05月19日

我有一个数据集，它有一个列(‘Facilities’)，它的对象数据类型和几个缺失值，以及一个不带空格的字符串值.如下所示:

如何为它们增添空间？我try 了下面的一些代码，但不起作用:

X['Restaurant'] = X['facilities'].apply(lambda x: 1 if 'Restaurant' in x else 0)
X['BAR'] = X['facilities'].apply(lambda x: 1 if 'BAR' in x else 0)
X['SwimmingPools'] = X['facilities'].apply(lambda x: 1 if 'SwimmingPools' in x else 0)

df3 = X['facilities'].str.split(n=1, expand=True)
df3.columns = ['STATUS_ID{}'.format(x+1) for x in df3.columns]

推荐答案

您可以使用re.split将单词拆分成一个列表，然后使用空格作为分隔符对列表进行.join:

import pandas as pd
import re

df = pd.DataFrame({"facilities":["GymrestaurantbarInternetSwimmingPools",
                                "Poolrestaurantgyminternetbar",
                                "BARswimmingPoolsInternetgym"]})

#                               facilities
# 0  GymrestaurantbarInternetSwimmingPools
# 1           Poolrestaurantgyminternetbar
# 2            BARswimmingPoolsInternetgym

pattern = '(gym|restaurant|internet|swimmingpools|bar)' #Add all the words you want to separate by here 

df["facilities_cleaned"] = df.apply(lambda x: " ".join([word for word in re.split(pattern=pattern, string=x["facilities"].lower()) if len(word)>0]), axis=1)

#                               facilities                         facilities_cleaned
# 0  GymrestaurantbarInternetSwimmingPools  gym restaurant bar internet swimmingpools
# 1           Poolrestaurantgyminternetbar           pool restaurant gym internet bar
# 2            BARswimmingPoolsInternetgym             bar swimmingpools internet gym