遍历日期

时间:2018-07-11 09:07:01

标签: python pandas loops date datetime

我试图遍历我创建的某些日期,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 ['完成日期1付款日期']是带有付款日期的熊猫列,而fin ['第一付款金额:\ n(70%)']是带有付款金额的熊猫列。

我收到以下错误

  

TypeError:键入对象2018-09-30

哪里出问题了?

1 个答案:

答案 0 :(得分:0)

我建议将to_datetimedate转换为datetimes,然后将DataFrame.loc用于选择列:

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)

或按string s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]

或升级到最新版本的熊猫以与日期进行比较,请选中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