Python 如何将熊猫文件保存为csv

Python 如何将熊猫文件保存为csv,python,pandas,csv,Python,Pandas,Csv,我有下面的代码如何将df文件保存为csv,如何在另一个程序中获取它,我想提取每个行元素,是否可以在输入相应日期时提取每个元素填充 import numpy as np from datetime import datetime,timedelta import pandas as pd # make your data a frame df = pd.DataFrame([[2020, 713000], [ 2019, 703

我有下面的代码如何将df文件保存为csv,如何在另一个程序中获取它,我想提取每个行元素,是否可以在输入相应日期时提取每个元素填充

    import numpy as np
    from datetime import datetime,timedelta
    import pandas as pd

    # make your data a frame
    df = pd.DataFrame([[2020,    713000], 
    [    2019,    703000], 
    [    2018,    694000], 
    [    2017,    684000], 
    [    2016,    674000], 
    [    2015,    664000], 
    [    2014,    655000], 
    [    2013,    645000], 
    [    2012,    636000], 
    [    2011,    627000]], columns=['DateTime','pop'])

    # make DateTime column an datetime object
    df['DateTime'] = df['DateTime'].apply(lambda x: datetime(x,1,1))

    # create a time range for each day in your period
    time_range = np.arange(datetime(2011, 1,1), datetime(2021,1,1), timedelta(days=1))

   # make time_range a frame 
   af = pd.DataFrame(time_range, columns=['DateTime'])

   # merge both together (left join on column DateTime) and interpolate the gaps
   df = af.merge(df, on='DateTime', how='left').interpolate()

   print(df)
我得到输出

     DateTime            pop
    0    2011-01-01  627000.000000
    1    2011-01-02  627024.657534
    2    2011-01-03  627049.315068
    3    2011-01-04  627073.972603
    4    2011-01-05  627098.630137
    ...         ...            ...
    3648 2020-12-27  713000.000000
    3649 2020-12-28  713000.000000
    3650 2020-12-29  713000.000000
    3651 2020-12-30  713000.000000
    3652 2020-12-31  713000.000000

查看文档,以便您决定是否要使用任何其他控制参数。

df.to\u csv,仅在StackOverflow上搜索pandas csv就返回36080个结果。这是否回答了您的问题?
df.to_csv("my_csvfilename.csv")