循环日期

Question

我正在尝试循环浏览我创建的一些日期，但出现错误。这是代码：

q3_2018 = datetime.date(2018,9,30) 
q4_2018 = datetime.date(2018,12,31) 
q1_2019 = datetime.date(2019,3,31) 
q2_2019 = datetime.date(2018,6,30) 

dates = [q3_2018, q4_2018,q1_2019,q2_2019]
values = []


for d in dates:
    v =  fin[fin['Date of Completion 1 payment']<d]['1st payment amount:\n(70%)'].sum() 
    values.append(v)

其中 fin['Date of Completion 1 payment'] 是带有付款日期的 pandas 列，而 fin['1st payment amount:\n(70%)'] 是带有已支付金额的 pandas 列。

我收到以下错误

TypeError：类型对象 2018-09-30

哪里出错了？

Answer 1

我建议通过to_datetime 将dates 转换为datetimes，然后选择列使用DataFrame.loc：

dates = pd.to_datetime([q3_2018, q4_2018,q1_2019,q2_2019])
print (dates)
DatetimeIndex(['2018-09-30', '2018-12-31', '2019-03-31', '2018-06-30'], 
              dtype='datetime64[ns]', freq=None)

或通过strings 进行比较：

dates = pd.to_datetime([q3_2018, q4_2018,q1_2019,q2_2019]).strftime('%Y-%m-%d')
print (dates)
Index(['2018-09-30', '2018-12-31', '2019-03-31', '2018-06-30'], dtype='object')

或者：

dates = ['2018-09-30', '2018-12-31' '2019-03-31','2018-06-30']

---

values = []
for d in dates:
    v =  fin.loc[fin['Date of Completion 1 payment']<d, '1st payment amount:\n(70%)'].sum() 
    values.append(v)

列表理解解决方案：

values = [fin.loc[fin['Date of Completion 1 payment']<d, '1st payment amount:\n(70%)'].sum() 
          for d in dates]

---

或者升级到最新版本的熊猫来比较日期，检查here：

# 0.22.0... Silently coerce the datetime.date
>>> Series(pd.date_range('2017', periods=2)) == datetime.date(2017, 1, 1)
0     True
1    False
dtype: bool

# 0.23.0... Do not coerce the datetime.date
>>> Series(pd.date_range('2017', periods=2)) == datetime.date(2017, 1, 1)
0    False
1    False
dtype: bool

# 0.23.1... Coerce the datetime.date with a warning
>>> Series(pd.date_range('2017', periods=2)) == datetime.date(2017, 1, 1)
/bin/python:1: FutureWarning: Comparing Series of datetimes with 'datetime.date'.  Currently, the
'datetime.date' is coerced to a datetime. In the future pandas will
not coerce, and the values not compare equal to the 'datetime.date'.
To retain the current behavior, convert the 'datetime.date' to a
datetime with 'pd.Timestamp'.
  #!/bin/python3
0     True
1    False
dtype: bool