Question

我在当前周开始日期和日期周开始之间取得了不同，其中日期是日期列。这种差异以天为单位。当我试图定义函数将这几天转换为1周差异，2周差异等等。

我收到的错误是：

无法将类型'Timedelta'与'str'类型进行比较

请帮我解决这个问题。在定义函数时我担心我错了吗？这是定义函数的代码：

def check(diff):
    for d in final_data['diff']:
        if  ((d > '0 days') and (d <= '7 days')): 
            weekdiff = 'OneWeekDiff'
        elif ((d > '8 days') and (d <= '14 days')):
            weekdiff = 'TwoWeekDiff'
        else:       
            weekdiff = 'Current Week'
    return weekdiff

要找出两列之间的差异，我只需要这样减法： final_data ['diff'] = final_data ['CurrentWeekStartDay'] - final_data ['InvoiceWeekstartDay']

打印（final_data [ '差异'] 0 14天 1 14天 2 14天

Answer 1

您需要转换to_timedelta字符串0 days，7 days ...：

然后我稍微修改一下功能 - 删除循环和else。您可以apply对DataFrame列

进行操作

def check(d):
    weekdiff = 'Current Week'
    if  ((d > pd.to_timedelta('0 days')) and (d <= pd.to_timedelta('7 days'))): 
        weekdiff = 'OneWeekDiff'
    elif ((d > pd.to_timedelta('8 days')) and (d <= pd.to_timedelta('14 days'))):
        weekdiff = 'TwoWeekDiff'
    return weekdiff

print (final_data['diff'].apply(check))

样品：

final_data = pd.DataFrame({'b': {0: pd.Timestamp('2016-01-13 00:00:00'), 
                                 1: pd.Timestamp('2016-01-05 00:00:00'), 
                                 2: pd.Timestamp('2016-01-03 00:00:00')}, 
                            'a': {0: pd.Timestamp('2016-01-01 00:00:00'), 
                                  1: pd.Timestamp('2016-01-02 00:00:00'), 
                                  2: pd.Timestamp('2016-01-03 00:00:00')}, 
                            'diff': {0: pd.Timedelta('12 days 00:00:00'), 
                                     1: pd.Timedelta('3 days 00:00:00'), 
                                     2: pd.Timedelta('0 days 00:00:00')}})
print (final_data)
           a          b    diff
0 2016-01-01 2016-01-13 12 days
1 2016-01-02 2016-01-05  3 days
2 2016-01-03 2016-01-03  0 days

print (final_data['diff'].apply(check))
0     TwoWeekDiff
1     OneWeekDiff
2    Current Week
Name: diff, dtype: object

如何将以天为单位的timedelta转换为字符串？

1 个答案: