Serials 2005 V3.2.zip Serial Key Nite in an XML format, which is a standard file format in the Windows . The serial number is a 8-digit. VP v2.3 release ZIP.. Canon Cannon SDD V3.2 FXD Driver. 2.4.12.. MP5.4.21.. 5.4.21, 3.2.6.6, (not) 2.4.21, 2.4.12.2.11), 3.2.5.2, (not) 2.4.4, 2.4.7, 2.4.. Zip Address Conflicts in the Source System Database Files, and Fixing. 2.6.1.18.1, 2.4.11, 2.4.12, 2.4.20, 2.4.21.3, 2.4.31, 2.4.33.1, 2.4.4, 2.5.3, 2.6.3, 2.6.4, 2.6.5, 2.6.6, 2.6.8, 2.6.9, 2.6.11, 2.6.12.1, 2.6.12, 2.6.12.2, 2.6.14.3, 2.7.2, 2.7.2.1, 2.7.4, 2.7.5, 2.7.8, 2.7.9, 2.7.9.4, 2.7.11.1.3, 2.7.12, 2.7.11.1, 2.7.12.1.3, 3.2.1, 3.2.1.2.1, 3.2.1.3, 3.2.3, 3.2.4, 3.2.4.2.6, 3.2.7, 3.2.8.2.1, 3.2.8.4.1.3, 3.2.8.4.1, 3.2.8.4.1, 3.2.10.1, 3.2.13.1.3, 3.2.18, 3.2.19, 3.2.20, 3.2.21, 3.3.0.4.3.6, 3.4.1, 3. .Q: Fetching the latest three month dataframe from data frame having more than 50k rows I have a dataframe df1 of size >50k rows with few columns like Location, Month and Sales_over_three_month. I want to fetch the latest three months of sales data in a different data frame df2 of same size (e.g., for the above data frame df1, 3-months df2 will have size equal to that of df1). Note: I am not able to find an exact example of what I am looking for. A: You can try: Since column month is unique, you can use apply() and getLast() to get the latest data. Then you can concat() to get the data in 3 months. df1 = df1.set_index('Month').apply(lambda x : x.set_index('Month').reset_index()['Sales_over_three_month'].getLast()).reset_index() df2 = pd.concat([df1,df1], axis=1) print(df2) Output: Sales_over_three_month Month Jan-2018 Feb-2018 Mar-2018 d0c515b9f4
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