按列排序，只保留第1行，直到第1列中的下一个值

Question

I have a file with roughly 10m lines. Each line is most likely unique, but I'm sorting the file by column 1 then 2 then 3.

Column 1 = CODE
Column 2 = DATE
Column 3 = AMOUNT

I only want to keep the first line until the next date and so on. Below is an example of what I have and what I need the output to be.

Original:  
COL1   COL2         COL3  
ABA    2019-01-01   100  
ABA    2019-01-01   111  
ABA    2019-01-02   140  
ABA    2019-01-02   150  
ABA    2019-01-03   200  
ABA    2019-01-03   220  

Ouptut needed:  
COL1   COL2         COL3  
ABA    2019-01-01   100  
ABA    2019-01-02   140  
ABA    2019-01-03   200  

Anyone able to help me. Have tried

a.drop_duplicates(subset[data.columns[0],data.columns[1],data.columns[2]], keep='first')

Answer 1

尝试groupby然后先：

a.groupby([data.columns[0],data.columns[1]], as_index=False).first()

Answer 2

Your solution is almost correct. This version is a modified version:

>> a.drop_duplicates(subset = [a.columns[0],a.columns[1]], keep='first')

That produces:

    COL1    COL2        COL3
0   ABA     2019-01-01  100
2   ABA     2019-01-02  140
4   ABA     2019-01-03  200

Explaining the modifications:

1. subset is a named parameter, as you can see on the documentation of drop_duplicates ;
2. if column 3 can vary, it shouldn't be present on the subset parameter. The duplicate should consider the first 2 columns;
3. the names you used in the code are not consistent, naming a and data for apparently the same object;