我是pandas
的新手,试图在Pandas Dataframe中创建新列,并基于函数分配字符串值,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)
答案 0 :(得分:1)
问题出在您的if循环上。 or
之后进行比较的方式不正确。
书写or 'Residence Hall/Dormitory'
将始终为true
,因此,每次仅对第一个if
进行求值,而您在所有行中都得到Residential
。
代替此:
if c['Primary Property Type - Self Selected'] == 'Multifamily Housing' or 'Residence Hall/Dormitory':
执行以下操作:
if c['Primary Property Type - Self Selected'] == 'Multifamily Housing' or c['Primary Property Type - Self Selected'] == 'Residence Hall/Dormitory':
OR
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,您的代码就可以完成预期的工作。希望这很清楚。
答案 1 :(得分:1)
在熊猫中,如果存在矢量化的解决方案,则最好避免使用循环(适用于引擎盖下的循环),因为循环很慢。
我尝试重写您的代码-使用输出和值列表创建字典,使用值交换键并调用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 Type
0 Medical Office Institutional
1 Hotel Commercial
2 Residence Hall/Dormitory Residential
3 Manufacturing/Industrial Plant Industrial
4 House Other