Python - 从excel文件中读取时间时没有得到正确的日期时间

Question

我有一个 excel 文件，它有 3 列作为日期时间或日期或时间字段。我正在通过xlrd 包阅读它，我想我得到了 milliseconds 的时间，当我试图将它转换回日期时间时，我得到了错误的结果。

我也尝试将文件转换为csv。这也无济于事，而且我得到了我无法理解的奇怪的日期时间格式。

这是我尝试使用xlrd 格式的方法。我更喜欢使用带有.xlrs 扩展名的文件作为输入，因为否则我每次得到一个新的输入文件时都必须将excel 文件转换为.csv。

from xlrd import open_workbook
import os,pickle,datetime

def main(path, filename, absolute_path_organisation_structure):
    absolute_filepath = os.path.join(path,filename)

    wb = open_workbook(absolute_filepath)
    for sheet in wb.sheets():
        number_of_rows = sheet.nrows
        number_of_columns = sheet.ncols

        for row_index in xrange(1, sheet.nrows):
            row=[]
            for col_index in xrange(4,7): #4th and 6th columns are date fields
                row.append(sheet.cell(row_index, col_index).value)

            print(row)  #Relevant list formed with 4th, 5th and 6th columns
            print(datetime.datetime.fromtimestamp(float(row[0])).strftime('%Y-%m-%d %H:%M:%S'))


path = "C:\\Users\\***************\\NEW DATA"
MISfile  = "P2P_2015 - Copy.xlsx"
absolute_path_organisation_structure = "C:\\Users\\******************NEW DATA\\organisation.csv"
main(path, MISfile, absolute_path_organisation_structure)

结果：

[42011.46789351852, u'Registered', 42009.0]
1970-01-01 17:10:11
[42011.46789351852, u'Sent for CTG1 approval', 42010.0]
1970-01-01 17:10:11
[42011.46789351852, u'Sent back', 42010.0]
1970-01-01 17:10:11
[42011.46789351852, u'Registered', 42011.0]
1970-01-01 17:10:11
[42011.46789351852, u'Sent for CTG1 approval', 42011.0]
1970-01-01 17:10:11
[42011.46789351852, u'Sent for CTG2 approval', 42012.0]
1970-01-01 17:10:11
[42011.46789351852, u'CTG2 Approved', 42012.0]
1970-01-01 17:10:11
[42011.46789351852, u'Sent back', 42013.0]
1970-01-01 17:10:11
[42170.61667824074, u'Registered', 42144.0]
1970-01-01 17:12:50
[42170.61667824074, u'Registered', 42144.0]
1970-01-01 17:12:50
[42170.61667824074, u'Sent back', 42165.0]
1970-01-01 17:12:50
[42170.61667824074, u'Sent back', 42165.0]
1970-01-01 17:12:50
[42170.61667824074, u'Registered', 42170.0]
1970-01-01 17:12:50
[42170.61667824074, u'Registered', 42170.0]
1970-01-01 17:12:50

实际输入文件：（从excel复制）

1/7/2015 11:13  Registered  1/5/2015 0:00
1/7/2015 11:13  Sent for CTG1 approval  1/6/2015 0:00
1/7/2015 11:13  Sent back   1/6/2015 0:00
1/7/2015 11:13  Registered  1/7/2015 0:00
1/7/2015 11:13  Sent for CTG1 approval  1/7/2015 0:00
1/7/2015 11:13  Sent for CTG2 approval  1/8/2015 0:00
1/7/2015 11:13  CTG2 Approved   1/8/2015 0:00
1/7/2015 11:13  Sent back   1/9/2015 0:00
6/15/2015 14:48 Registered  5/20/2015 0:00
6/15/2015 14:48 Registered  5/20/2015 0:00
6/15/2015 14:48 Sent back   6/10/2015 0:00
6/15/2015 14:48 Sent back   6/10/2015 0:00
6/15/2015 14:48 Registered  6/15/2015 0:00
6/15/2015 14:48 Registered  6/15/2015 0:00

为什么我无法正确读取日期？为什么它们不简单地作为字符串出现以便我可以轻松地转换它们？

Answer 1

如果要读取的 Excel 文件是表格可以简单直接地使用pandas.read_excel。 使用pandas.to_datetime 转换日期后

from __future__ import absolute_import, division, print_function
import os
import pandas as pd

def main(path, filename, absolute_path_organisation_structure):
    absolute_filepath = os.path.join(path,filename)
    #Relevant list formed with 4th, 5th and 6th columns
    df = pd.read_excel(absolute_filepath, header=None, parse_cols=[4,5,6])
    # Transform column 0 and 2 to datetime
    df[0] = pd.to_datetime(df[0])
    df[2] = pd.to_datetime(df[2])
    print(df)

path = "C:\\Users\\***************\\NEW DATA"
MISfile  = "P2P_2015 - Copy.xlsx"
main(path, MISfile,None)

Answer 2

xldate_as_tuple(xldate, datemode) [#]

将 Excel 数字（假定表示日期、日期时间或时间）转换为适合提供给 datetime 或 mx.DateTime 构造函数的元组。

来源：http://www.lexicon.net/sjmachin/xlrd.html#xlrd.xldate_as_tuple-function

用法示例：How to use ``xlrd.xldate_as_tuple()``

Answer 3

问题是您将 Excel 日期时间值解释为 UNIX 时间戳，而它们不是同一事物。要查找的警告标志是结果值都接近 UNIX 纪元 (1970-01-01)。

您可以使用in this answer 描述的方法将 Excel 日期时间转换为 UNIX。

Windows/Mac Excel 2011

Unix Timestamp = (Excel Timestamp - 25569) * 86400

Mac Excel 2007

Unix Timestamp = (Excel Timestamp - 24107) * 86400

如果你应用这个转换，你应该得到正确的输出：

timestamp = (float(row[0]) - 25569) * 86400
datetime.datetime.fromtimestamp(timestamp).strftime('%Y-%m-%d %H:%M:%S')