我有如下資料框
cust_id,purchase_date
1,10/01/1998
1,10/12/1999
2,13/05/2016
3,14/02/2018
3,15/03/2019
我想做以下
a）以文本格式顯示輸出，5 years and 9 months而不是5.93244等。
我嘗試了以下
from datetime import timedelta
df['purchase_date'] = pd.to_datetime(df['purchase_date'])
gb = df_new.groupby(['unique_key'])
df_cust_age = gb['purchase_date'].agg(min_date=np.min, max_date=np.max).reset_index()
df_cust_age['diff_in_days'] = df_cust_age['max_date'] - df_cust_age['min_date']
df_cust_age['years_diff'] = df_cust_age['diff_in_days']/timedelta(days=365)
但上面的代碼給出了輸出decimal numbers。
我希望我的輸出如下
cust_id,years_diff
1, 1 years and 11 months and 0 day
2, 0 years
3, 1 year and 1 month and 1 day
uj5u.com熱心網友回復：
如果可能'default'，使用此自定義函式創建 30 天的月份：
#https://stackoverflow.com/a/13756038/2901002
def td_format(td_object):
seconds = int(td_object.total_seconds())
periods = [
('year', 60*60*24*365),
('month', 60*60*24*30),
('day', 60*60*24),
('hour', 60*60),
('minute', 60),
('second', 1)
]
strings=[]
for period_name, period_seconds in periods:
if seconds > period_seconds:
period_value , seconds = divmod(seconds, period_seconds)
has_s = 's' if period_value > 1 else ''
strings.append("%s %s%s" % (period_value, period_name, has_s))
return ", ".join(strings) if len(strings) > 0 else '0 year'
df_cust_age['years_diff'] = df_cust_age['diff_in_days'].apply(td_format)
print (df_cust_age)
cust_id min_date max_date diff_in_days years_diff
0 1 1998-10-01 1999-10-12 376 days 1 year, 11 days
1 2 2016-05-13 2016-05-13 0 days 0 year
2 3 2018-02-14 2019-03-15 394 days 1 year, 29 days
uj5u.com熱心網友回復：
import datetime as dt
DATEFUL = ['{}-{}-15'.format(year, int(month) if int(month) >= 10 else '0{}'.format(int(month))) for year, month in zip(['1998','2009','2016','2018','2019'], ['10', '12', '05', '02', '03'])]
# Testing
df = pd.DataFrame({'lst': [10, 12, 5, 2, 3],
'dates': DATEFUL})
# Expected output function
def fratedelta(duration):
days, seconds = duration.days, duration.seconds
hours = seconds // 3600
minutes = (seconds % 3600) // 60
seconds = (seconds % 60)
if days > 365*24*60*60:
years, days = days//365, days%365
return '{} years and {} days'.format(years, days)
else:
return '{} days and {} hours'.format(days, hours)
today=dt.date.today()
df['years_diff'] = df.date.apply(lambda x:
fratedelta(today - dt.datetime.strptime(x,'%Y-%m-%d')))
print(df)
uj5u.com熱心網友回復：
from io import StringIO
import pandas as pd
from dateutil.relativedelta import relativedelta as RD
string_data = '''unique_key,purchase_date
1,10/01/1998
1,10/12/1999
2,13/05/2016
3,14/02/2018
3,15/03/2019'''
## Custom functions
diff_obj = lambda d1,d2:RD(d1, d2) if d1>d2 else RD(d2, d1)
date_tuple = lambda diff:(diff.years,diff.months,diff.days)
pipeline = lambda row:date_tuple(diff_obj(row['min_date'],row['max_date']))
def string_format(date_tuple):
final_string = []
for val,name in zip(date_tuple,['years','months','day']):
if val:
final_string.append(f'{val} {name}')
return ' and '.join(final_string) if final_string else '0 years'
## Custom functions
df = pd.read_csv(StringIO(string_data))
df['purchase_date'] = pd.to_datetime(df['purchase_date'],format='%d/%m/%Y')
gb = df.groupby(['unique_key'])
df_cust_age = gb['purchase_date'].agg(min_date=np.min, max_date=np.max).reset_index()
df_cust_age['years_diff'] = df_cust_age.apply(pipeline,axis=1).apply(string_format)
print(df_cust_age)
unique_key min_date max_date years_diff
0 1 1998-01-10 1999-12-10 1 years and 11 months
1 2 2016-05-13 2016-05-13 0 years
2 3 2018-02-14 2019-03-15 1 years and 1 months and 1 day
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