在pandas df中创建新列并有条件地分配字符串值

我是新手pandas,试图在 Pandas Dataframe 中创建一个新列,并根据函数分配一个字符串值,但结果仅向所有 5,000 列输出 1 个值('住宅)。知道我的代码有什么问题吗?谢谢


def programType(c):

if c['Primary Property Type - Self Selected'] == 'Multifamily Housing' or 'Residence Hall/Dormitory':

    return 'Residential'


elif c['Primary Property Type - Self Selected'] == 'Bank Branch' or 'Hotel' or 'Financial Office' \

or 'Retail Store' or 'Distribution Center' or 'Non-Refrigerated Warehouse' or 'Fitness Center/Health Club/Gym' \

or 'Mixed Use Property' or 'Self-Storage Facility' or 'Wholesale Club/Supercenter' or 'Supermarket/Grocery Store':

    return 'Commercial'  


elif c['Primary Property Type - Self Selected'] == 'Senior Care Community' or 'K-12 School' or 'College/University' \

or 'Worship Facility' or 'Medical Office' or 'Hospital (General Medical & Surgical)':

    return 'Institutional'


elif c['Primary Property Type - Self Selected'] == 'Manufacturing/Industrial Plant': 

    return 'Industrial'


else:

    return 'Other'

新列称为“程序类型”


datav3['Program Type'] = datav3.apply(programType, axis=1)


POPMUISE
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2回答

慕容3067478

如果存在矢量化解决方案,在 Pandas 中最好避免循环(应用是引擎盖下的循环),因为循环很慢。我尝试重写您的代码 - 创建带有输出和值列表的字典,用值交换键并调用map,最后为不匹配的值添加fillna:d = {'Residential' :['Multifamily Housing', 'Residence Hall/Dormitory'],     'Commercial' : ['Bank Branch', 'Hotel' , 'Financial Office' , 'Retail Store', 'Distribution Center',                    'Non-Refrigerated Warehouse', 'Fitness Center/Health Club/Gym', 'Mixed Use Property',                   'Self-Storage Facility', 'Wholesale Club/Supercenter', 'Supermarket/Grocery Store'],     'Institutional':['Senior Care Community', 'K-12 School', 'College/University', 'Worship Facility',                     'Medical Office', 'Hospital (General Medical & Surgical)'],     'Industrial':  ['Manufacturing/Industrial Plant'] }d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}print (d1){    'Multifamily Housing': 'Residential',    'Residence Hall/Dormitory': 'Residential',    'Bank Branch': 'Commercial',    'Hotel': 'Commercial',    'Financial Office': 'Commercial',    'Retail Store': 'Commercial',    'Distribution Center': 'Commercial',    'Non-Refrigerated Warehouse': 'Commercial',    'Fitness Center/Health Club/Gym': 'Commercial',    'Mixed Use Property': 'Commercial',    'Self-Storage Facility': 'Commercial',    'Wholesale Club/Supercenter': 'Commercial',    'Supermarket/Grocery Store': 'Commercial',    'Senior Care Community': 'Institutional',    'K-12 School': 'Institutional',    'College/University': 'Institutional',    'Worship Facility': 'Institutional',    'Medical Office': 'Institutional',    'Hospital (General Medical & Surgical)': 'Institutional',    'Manufacturing/Industrial Plant': 'Industrial'}datav3 = pd.DataFrame({'Program':['Medical Office','Hotel',                                       'Residence Hall/Dormitory',                                       'Manufacturing/Industrial Plant','House']})datav3['Program Type'] = datav3['Program'].map(d1).fillna('Other')print (datav3)                          Program   Program Type0                  Medical Office  Institutional1                           Hotel     Commercial2        Residence Hall/Dormitory    Residential3  Manufacturing/Industrial Plant     Industrial4                           House          Other

素胚勾勒不出你

问题在于您的 if 循环。您之后比较的方式or不正确。写作or 'Residence Hall/Dormitory' 将永远是true,因此,if每次只评估第一个,并且您进入Residential所有行。取而代之的是:if c['Primary Property Type - Self Selected'] == 'Multifamily Housing' or 'Residence Hall/Dormitory':做这个:if c['Primary Property Type - Self Selected'] == 'Multifamily Housing' or c['Primary Property Type - Self Selected'] == 'Residence Hall/Dormitory':或者if any([c['Primary Property Type - Self Selected'] == 'Multifamily Housing', c['Primary Property Type - Self Selected'] == 'Residence Hall/Dormitory']):只需进行上述更改,您的代码就应该符合预期。希望这很清楚。
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