Pandas assign inplace example

inplace_example.py

# tl;dr: you can't assign in place because inplace returns None

import pandas as pd
dat = pd.util.testing.makeMixedDataFrame()
print(dat)

#      A    B     C          D
# 0  0.0  0.0  foo1 2009-01-01
# 1  1.0  1.0  foo2 2009-01-02
# 2  2.0  0.0  foo3 2009-01-05
# 3  3.0  1.0  foo4 2009-01-06
# 4  4.0  0.0  foo5 2009-01-07

# individual assignment with [ ]
dat1 = dat.copy()
dat1["new_col_1"] = dat1["A"] + dat1["B"] ## create new column
dat1["new_col_2"] = dat1["new_col_1"]*10  ## use new column
dat1 = dat1.loc[dat1["new_col_1"] >= 2]   ## filter on existing column
print(dat1)

#      A    B     C          D  new_col_1  new_col_2
# 1  1.0  1.0  foo2 2009-01-02        2.0       20.0
# 2  2.0  0.0  foo3 2009-01-05        2.0       20.0
# 3  3.0  1.0  foo4 2009-01-06        4.0       40.0
# 4  4.0  0.0  foo5 2009-01-07        4.0       40.0

# assign back to dataframe works just fine
dat2 = dat.copy()
dat2 = dat2.assign(a = 3,
                   b = lambda x: x["a"] * 10) ## using new column needs lambda notation
print(dat2)

#      A    B     C          D  a   b
# 0  0.0  0.0  foo1 2009-01-01  3  30
# 1  1.0  1.0  foo2 2009-01-02  3  30
# 2  2.0  0.0  foo3 2009-01-05  3  30
# 3  3.0  1.0  foo4 2009-01-06  3  30
# 4  4.0  0.0  foo5 2009-01-07  3  30

# case for "inplace"
dat2 = dat.copy()
dat2 = dat2.assign(new_col_1 = lambda x: x["A"] + x["B"],
            new_col_2 = lambda x: x["new_col_1"]*10
).loc[dat2["new_col_1"] >= 2]
dat2

# KeyError: 'new_col_1' ## from the .loc call

# you would need to re-write the above as such
dat2 = dat.copy()
dat2 = dat2.assign(new_col_1 = lambda x: x["A"] + x["B"],
            new_col_2 = lambda x: x["new_col_1"]*10
)
dat2 = dat2.loc[dat2["new_col_1"] >= 2]
print(dat2)

#      A    B     C          D  new_col_1  new_col_2
# 1  1.0  1.0  foo2 2009-01-02        2.0       20.0
# 2  2.0  0.0  foo3 2009-01-05        2.0       20.0
# 3  3.0  1.0  foo4 2009-01-06        4.0       40.0
# 4  4.0  0.0  foo5 2009-01-07        4.0       40.0


# what it would look like with an "inplace_" parameter
# "inplace_" instead of "inplace" just incase there's a column named "inpalce"
# this woudln't work because normally inplace returns None
dat2 = dat.copy()
dat2 = (dat2
        .assign(new_col_1 = lambda x: x["A"] + x["B"],
                new_col_2 = lambda x: x["new_col_1"]*10,
                inplace_ = True
                )
        .loc[dat2["new_col_1"] >= 2]
print(dat2)

# would expect the output to be the same as dat1 above:
#      A    B     C          D  new_col_1  new_col_2
# 1  1.0  1.0  foo2 2009-01-02        2.0       20.0
# 2  2.0  0.0  foo3 2009-01-05        2.0       20.0
# 3  3.0  1.0  foo4 2009-01-06        4.0       40.0
# 4  4.0  0.0  foo5 2009-01-07        4.0       40.0


# Other methods that dont require direct reference would work
# below i show `.drop()`
dat3 = dat.copy()
dat3 = (dat3
        .assign(new_col_1 = lambda x: x["A"] + x["B"],
                new_col_2 = lambda x: x["new_col_1"]*10)
        .drop(columns=["A", "B", "C"])
)

print(dat3)

#            D  new_col_1  new_col_2
# 0 2009-01-01        0.0        0.0
# 1 2009-01-02        2.0       20.0
# 2 2009-01-05        2.0       20.0
# 3 2009-01-06        4.0       40.0
# 4 2009-01-07        4.0       40.0

# .loc will stil fail in the example, even if you put inplace within drop call
# this is because inplace retuns None

# instead you would fix all of this by using query
dat3 = dat.copy()
dat3 = (dat3
        .assign(new_col_1 = lambda x: x["A"] + x["B"],
                new_col_2 = lambda x: x["new_col_1"]*10)
        .drop(columns=["A", "B", "C"])
        .query('new_col_1 >= 2')
)

print(dat3)

#            D  new_col_1  new_col_2
# 1 2009-01-02        2.0       20.0
# 2 2009-01-05        2.0       20.0
# 3 2009-01-06        4.0       40.0
# 4 2009-01-07        4.0       40.0