NOD32 V2.51.2 – NOD32 FiX V2.1 Web2k2 Serial Key _VERIFIED_

NOD32 V2.51.2 – NOD32 FiX V2.1 Web2k2 Serial Key _VERIFIED_



 
 
 
 
 
 
 

NOD32 V2.51.2 – NOD32 FiX V2.1 Web2k2 Serial Key

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Download latest version of best antivirus software on PC (or) Laptop from here. It works very fast.Q:

How to get unique row based on specific column and not by sorting all the rows?

I have 2 columns in dataframe, one is number and the other is flag with Boolean type.
def flag_column(df):
return df[‘flag’].where(df[‘flag’].isin([“True”, “True”, “False”, “False”]))

df1 = pd.DataFrame(data={“flag”:[True,True,False,True,False],
“number”:[1,1,1,1,1]})
df1

But it returns the unique data based on the order of each row. How do I get the unique data based on flag column and not on the order of each row?
Please note that I do not want to get the data sorted first, then unique. I just want to see the unique rows based on flag column.
The result I want is:
1 True
1 True

A:

You can use the | bitwise OR operator, here, to combine flags:
>>> import numpy as np
>>> df[“flag”].values = np.array([True,True,False,True,True])
>>> df[df[“flag”].apply(lambda x: x|1)]
number flag
0 1 True
1 1 True
2 1 False
3 1 True

mask = df[“flag”].apply(lambda x: x|1)
df[df[“flag”].isin([“True”, “True”, “False”, “False”])]

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