True is 1 in Python, and likewise False is 0^*:

>>> True == 1
True
>>> False == 0
True

You should be able to perform any operations you want on them by just treating them as though they were numbers, as they are numbers:

>>> issubclass(bool, int)
True
>>> True * 5
5

So to answer your question, no work necessary – you already have what you are looking for.

^{* Note I use is as an English word, not the Python keyword is – True will not be the same object as any random 1.}

You also can do this directly on Frames

In [104]: df = DataFrame(dict(A = True, B = False),index=range(3))

In [105]: df
Out[105]: 
      A      B
0  True  False
1  True  False
2  True  False

In [106]: df.dtypes
Out[106]: 
A    bool
B    bool
dtype: object

In [107]: df.astype(int)
Out[107]: 
   A  B
0  1  0
1  1  0
2  1  0

In [108]: df.astype(int).dtypes
Out[108]: 
A    int64
B    int64
dtype: object

Just multiply your Dataframe by 1 (int)

[1]: data = pd.DataFrame([[True, False, True], [False, False, True]])
[2]: print data
          0      1     2
     0   True  False  True
     1   False False  True

[3]: print data*1
         0  1  2
     0   1  0  1
     1   0  0  1

You can use a transformation for your data frame:

df = pd.DataFrame(my_data condition)

transforming True/False in 1/0

df = df*1

Use Series.view for convert boolean to integers:

df["somecolumn"] = df["somecolumn"].view('i1')

I had to map FAKE/REAL to 0/1 but couldn’t find proper answer.

Please find below how to map column name ‘type’ which has values FAKE/REAL to 0/1
(Note: similar can be applied to any column name and values)

df.loc[df['type'] == 'FAKE', 'type'] = 0
df.loc[df['type'] == 'REAL', 'type'] = 1

This question specifically mentions a single column, so the currently accepted answer works. However, it doesn’t generalize to multiple columns. For those interested in a general solution, use the following:

df.replace({False: 0, True: 1}, inplace=True)

This works for a DataFrame that contains columns of many different types, regardless of how many are boolean.

This is a reproducible example based on some of the existing answers:

import pandas as pd


def bool_to_int(s: pd.Series) -> pd.Series:
    """Convert the boolean to binary representation, maintain NaN values."""
    return s.replace({True: 1, False: 0})


# generate a random dataframe
df = pd.DataFrame({"a": range(10), "b": range(10, 0, -1)}).assign(
    a_bool=lambda df: df["a"] > 5,
    b_bool=lambda df: df["b"] % 2 == 0,
)

# select all bool columns (or specify which cols to use)
bool_cols = [c for c, d in df.dtypes.items() if d == "bool"]

# apply the new coding to a new dataframe (or can replace the existing one)
df_new = df.assign(**{c: lambda df: df[c].pipe(bool_to_int) for c in bool_cols})

Tries and tested:

df[col] = df[col].map({'True': 1,'False' :0 })

If there are more than one columns with True/False, use the following.

for col in bool_cols:
    df[col] = df[col].map({'True': 1,'False' :0 })

@AMC wrote this in a comment

If the column is of the type object, and for example you want to convert it to integer:

df["somecolumn"] = df["somecolumn"].astype(bool).astype(int)