Factorize a column of strings in pandas

As the question says, I have a data frame df_original which is quite large but looks like:

        ID    Count   Column 2   Column 3  Column 4
RowX    1      234.     255.       yes.      452
RowY    1      123.     135.       no.       342
RowW    1      234.     235.       yes.      645
RowJ    1      123.     115.       no.       342
RowA    1      234.     285.       yes.      233
RowR    1      123.     165.       no.       342
RowX    2      234.     255.       yes.      234
RowY    2      123.     135.       yes.      342
RowW    2      234.     235.       yes.      233
RowJ    2      123.     115.       yes.      342
RowA    2      234.     285.       yes.      312
RowR    2      123.     165.       no.       342
.
.
.
RowX    1233   234.     255.       yes.      133
RowY    1233   123.     135.       no.       342
RowW    1233   234.     235.       no.       253
RowJ    1233   123.     115.       yes.      342
RowA    1233   234.     285.       yes.      645
RowR    1233   123.     165.       no.       342

I am trying to get rid of the text data and replace it with a predefined numerical equivalent. For example, in this case, I'd like to replace Column3's yes or no values with 1 or 0 respectively. Is there a way to do this without me having to manually go in and alter the values?

v

RowX    yes
RowY     no
RowW    yes
RowJ     no
RowA    yes
RowR     no
RowX    yes
RowY    yes
RowW    yes
RowJ    yes
RowA    yes
RowR     no
Name: Column 3, dtype: object

`pd.factorize`

1 - pd.factorize(v)[0]
array([1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0])

`np.where`

np.where(v == 'yes', 1, 0)
array([1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0])

`pd.Categorical`/`astype('category')`

pd.Categorical(v).codes
array([1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0], dtype=int8)

v.astype('category').cat.codes

RowX    1
RowY    0
RowW    1
RowJ    0
RowA    1
RowR    0
RowX    1
RowY    1
RowW    1
RowJ    1
RowA    1
RowR    0
dtype: int8

`pd.Series.replace`

v.replace({'yes' : 1, 'no' : 0})

RowX    1
RowY    0
RowW    1
RowJ    0
RowA    1
RowR    0
RowX    1
RowY    1
RowW    1
RowJ    1
RowA    1
RowR    0
Name: Column 3, dtype: int64

A fun, generalised version of the above:

v.replace({r'^(?!yes).*$' : 0}, regex=True).astype(bool).astype(int)

RowX    1
RowY    0
RowW    1
RowJ    0
RowA    1
RowR    0
RowX    1
RowY    1
RowW    1
RowJ    1
RowA    1
RowR    0
Name: Column 3, dtype: int64

Anything that is not "yes" is 0.