I was replacing values in columns and noticed that if use mask on all the dataframe, it will produce expected results, but if I used it against selected columns with .loc, it won't change any value.
Can you explain why and tell if it is expected result?
You can try with a dataframe dt, containing 0 in columns:
dt = pd.DataFrame(np.random.randint(0,3,size=(10, 3)), columns=list('ABC'))
dt.mask(lambda x: x == 0, np.nan, inplace=True)
# will replace all zeros to nan, OK.
But:
dt = pd.DataFrame(np.random.randint(0,3,size=(10, 3)), columns=list('ABC'))
columns = list('BC')
dt.loc[:, columns].mask(lambda x: x == 0, np.nan, inplace=True)
# won't cange anything. I excpet B, C columns to have values replaced
CodePudding user response：
i guess it's because the DataFrame.loc property is just giving access to a slice of your dataframe and you are masking a copy of the dataframe so it doesn't affect the data.
you can try this instead:
dt[columns] = dt[columns].mask(dt[columns] == 0)
CodePudding user response：
The loc functions returns a copy of the dataframe. On this copy you are applying the mask function that perform the operation in place on the data. You can't do this on a one-liner, otherwise the memory copy remains inaccessible. To get access to that memory area you have to split the code into 2 lines, to get a reference to that memory area:
tmp = dt.loc[:, columns]
tmp.mask(tmp[columns] == 0, np.nan, inplace=True)
and then you can go and update the dataframe:
dt[columns] = tmp
Not using the inplace update of the mask function, on the other hand, you can do everything with one line of code
dt[columns] = dt.loc[:, columns].mask(dt[columns] == 0, np.nan, inplace=False)
Extra: If you want to better understand the use of the inplace method in pandas, I recommend you read these posts: