按年份过滤行； AttributeError：只能将 .dt 访问器与 datetimelike 值一起使用答案

【问题标题】：Filtering rows by years; AttributeError: Can only use .dt accessor with datetimelike values按年份过滤行； AttributeError：只能将 .dt 访问器与 datetimelike 值一起使用
【发布时间】：2018-08-30 00:09:45
【问题描述】：

此代码用于显示哪些交货延迟，打印出与其关联的“材料”编号，并显示交货延迟了多少天。我现在的问题在于尝试过滤数据集以仅读取指定的时间范围；在我的以下代码中，我尝试过滤从 2017 年到 2018 年的数据，但是我收到一个错误（在代码块下方列出）。如何过滤行以仅显示指定的时间范围，同时进行相同的分析：即查看哪些材料零件号延迟交货并查看延迟了多少天（没有遇到错误）

import pandas as pd
from datetime import datetime
from datetime import timedelta



df = pd.read_csv('otd.csv')

diff_delivery_date = []
date_format = '%m/%d/%Y'
df2 = df[(df['Delivery Date'].dt.year >= 2017) & (df['Delivery Date'].dt.year <= 2018)]



for x,y,z in zip(df2['Material'], df2['Delivery Date'], df2['source desired delivery date']):
    actual_deliv_date = datetime.strptime(y, date_format)
    supposed_deliv_date = datetime.strptime(z, date_format)
    diff_deliv_date = supposed_deliv_date - actual_deliv_date
    diff_delivery_date.append(diff_deliv_date)

df['Diff Deliv Date'] = diff_delivery_date

print(df2)

完全错误：

Traceback (most recent call last):
  File "C:\Users\khalha\eclipse-workspace\Python\Heyy\Code.py", line 13, in <module>
    df2 = df[(df['Delivery Date'].dt.year >= 2017) & (df['Delivery Date'].dt.year <= 2018)]
  File "C:\Users\khalha\AppData\Local\Programs\Python\Python37\lib\site-packages\pandas\core\generic.py", line 4372, in __getattr__
    return object.__getattribute__(self, name)
  File "C:\Users\khalha\AppData\Local\Programs\Python\Python37\lib\site-packages\pandas\core\accessor.py", line 133, in __get__
    accessor_obj = self._accessor(obj)
  File "C:\Users\khalha\AppData\Local\Programs\Python\Python37\lib\site-packages\pandas\core\indexes\accessors.py", line 325, in __new__
    raise AttributeError("Can only use .dt accessor with datetimelike "
AttributeError: Can only use .dt accessor with datetimelike values

虚拟 csv： Image of csv file

Material    Delivery Date   source desired delivery date
3334678 12/31/2014  12/31/2014
233433  12/31/2014  12/31/2014
3434343 1/5/2015    1/5/2015
3334567 1/5/2015    1/5/2015
546456  2/11/2015   2/11/2015
221295  4/10/2015   4/10/2015

示例数据框：

Deliveryvalue = df2['11/31/2014', '11/31/2017', '11/31/2018']
Desiredvalue = df2['12/31/2014', '12/21/2017', '12/11/2018']

【问题讨论】：

您能否提供一个可以使用的DataFrame示例（不是图像）？
@DanielMesejo，好的，我应该如何提供这个数据框样本，我对 Stackoverflow 的新手表示歉意。
类似 pd.DataFrame(data=[[3334678, '12/31/2014', '12/31/2014']], columns=['material', 'delivery-date' , 'source-desired-delivery'])，其中的值反映了数据的真实值和类型。
“otd.csv”是用逗号还是空格分隔的？
@DanielMesejo，我相信它的逗号分隔，当我打印某种列表时，有逗号

标签： python pandas csv datetime

【解决方案1】：

这个答案我假设您的数据具有以下格式：

Material,Delivery Date,source desired delivery date
3334678,12/31/2017,12/31/2017
233433,12/31/2017,12/31/2017
3434343,1/5/2017,1/5/2017
3334567,1/5/2017,1/5/2017
546456,2/11/2017,2/11/2017
221295,4/10/2017,4/10/2017

所以，假设你可以这样做：

import pandas as pd

df = pd.read_csv('odt.csv')

df['Delivery Date'] = pd.to_datetime(df['Delivery Date'], format='%m/%d/%Y')
df['source desired delivery date'] = pd.to_datetime(df['source desired delivery date'], format='%m/%d/%Y')

df2 = df[(df['Delivery Date'].dt.year >= 2017) & (df['Delivery Date'].dt.year <= 2018)]
df2['Diff Deliv Date'] = df2['Delivery Date'] - df2['source desired delivery date']

print(df2)

输出

   Material Delivery Date source desired delivery date Diff Deliv Date
0   3334678    2017-12-31                   2017-12-31          0 days
1    233433    2017-12-31                   2017-12-31          0 days
2   3434343    2017-01-05                   2017-01-05          0 days
3   3334567    2017-01-05                   2017-01-05          0 days
4    546456    2017-02-11                   2017-02-11          0 days
5    221295    2017-04-10                   2017-04-10          0 days

备注

加载数据后列的类型如下：

Material                         int64
Delivery Date                   object
source desired delivery date    object

你可以检查你的是否是那些。然后您需要将'Delivery Date' 和'source desired delivery date' 转换为日期时间，这是在：

df['Delivery Date'] = pd.to_datetime(df['Delivery Date'], format='%m/%d/%Y')
df['source desired delivery date'] = pd.to_datetime(df['source desired delivery date'], format='%m/%d/%Y')

然后简单地过滤数据并计算差异。我也改变了：

df['Diff Deliv Date'] = diff_delivery_date

给df2，而不是你的代码最后打印df2。

【讨论】：