Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

one hot encoding for frequent values only

I am looking to do one hot encoding to a column, but only for those that are very frequent. All that are below a threshold T will be put in their own category.

My strategy was to create a dictionary of "name" -> "frequency". Then convert the frequency to a string. If the string is uncommon, it should be replaced with some descriptive string. Preferably, I want to have two regions/thresholds: "less_common" and "rare" or something like that.

Here is my current attempt. I have it broken into lines just for debugging fyi. The 3rd line does not work. I am using conda with Python 3.6,

tmp = df["name"].groupby(df["name"])
tmp = tmp.agg(['count'])
tmp['count'] = tmp["count"].apply(lambda x: "Uncommon" if tmp["count"] < 1000.0 else str(x) )
labelDict = tmp.to_dict()
#some code?
df[columnName].replace(labelDict, inplace=True)
pd.get_dummies(df, columns=['name'])

The error:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Some example input (there are other columns): name = a, a, a, a, b, b, b, c, c, d

This becomes

name | count
a | 4
b | 3
c | 2
d | 1

Let's say T is =<2
dict:
a->4, b->3, c->"Uncommon", d->"Uncommon"

Remap dict to use the original values if name is numeric:
a->"a", b->"b", c->"Uncommon", d->"Uncommon"

As one hot:
date | id | name_a | name_b | name_Uncommon 
...  | ...|  1     | 0      | 0
...

Current lib versions:

['alabaster==0.7.10', 'anaconda-client==1.6.3', 'anaconda-navigator==1.6.2', 'anaconda-project==0.6.0', 'appnope==0.1.0', 'appscript==1.0.1', 'asn1crypto==0.22.0', 'astroid==1.4.9', 'astropy==1.3.2', 'babel==2.4.0', 'backports.shutil-get-terminal-size==1.0.0', 'beautifulsoup4==4.6.0', 'bitarray==0.8.1', 'blaze==0.10.1', 'bleach==1.5.0', 'bokeh==0.12.5', 'boto==2.46.1', 'bottleneck==1.2.1', 'branca==0.2.0', 'cffi==1.10.0', 'chardet==3.0.3', 'chest==0.2.3', 'click==6.7', 'cloudpickle==0.2.2', 'clyent==1.2.2', 'colorama==0.3.9', 'conda==4.3.21', 'configobj==5.0.6', 'contextlib2==0.5.5', 'cryptography==1.8.1', 'cycler==0.10.0', 'cython==0.25.2', 'cytoolz==0.8.2', 'dask==0.14.3', 'datashape==0.5.4', 'decorator==4.0.11', 'dill==0.2.6', 'distributed==1.16.3', 'docutils==0.13.1', 'entrypoints==0.2.2', 'et-xmlfile==1.0.1', 'fastcache==1.0.2', 'flask-cors==3.0.2', 'flask==0.12.2', 'folium==0.3.0', 'gevent==1.2.1', 'greenlet==0.4.12', 'h5py==2.7.0', 'heapdict==1.0.0', 'html5lib==0.9999999', 'idna==2.5', 'imagesize==0.7.1', 'ipykernel==4.6.1', 'ipython-genutils==0.2.0', 'ipython==5.3.0', 'ipywidgets==6.0.0', 'isort==4.2.5', 'itsdangerous==0.24', 'jdcal==1.3', 'jedi==0.10.2', 'jinja2==2.9.6', 'jsonschema==2.6.0', 'jupyter-client==5.0.1', 'jupyter-console==5.1.0', 'jupyter-core==4.3.0', 'jupyter==1.0.0', 'keras==2.0.4', 'lazy-object-proxy==1.2.2', 'llvmlite==0.18.0', 'locket==0.2.0', 'lxml==3.7.3', 'mako==1.0.6', 'markdown==2.2.0', 'markupsafe==0.23', 'matplotlib==2.0.2', 'mistune==0.7.4', 'mpmath==0.19', 'msgpack-python==0.4.8', 'multipledispatch==0.4.9', 'navigator-updater==0.1.0', 'nbconvert==5.1.1', 'nbformat==4.3.0', 'networkx==1.11', 'nltk==3.2.3', 'nose==1.3.7', 'notebook==5.0.0', 'numba==0.33.0', 'numexpr==2.6.2', 'numpy==1.12.1', 'numpydoc==0.6.0', 'odo==0.5.0', 'olefile==0.44', 'openpyxl==2.4.7', 'packaging==16.8', 'pandas==0.20.1', 'pandocfilters==1.4.1', 'partd==0.3.8', 'pathlib2==2.2.1', 'patsy==0.4.1', 'pep8==1.7.0', 'pexpect==4.2.1', 'pickleshare==0.7.4', 'pillow==4.1.1', 'pip==9.0.1', 'ply==3.10', 'prompt-toolkit==1.0.14', 'protobuf==3.3.0', 'psutil==5.2.2', 'ptyprocess==0.5.1', 'py==1.4.33', 'pyasn1==0.2.3', 'pycosat==0.6.2', 'pycparser==2.17', 'pycrypto==2.6.1', 'pycurl==7.43.0', 'pyflakes==1.5.0', 'pygments==2.2.0', 'pygpu==0.6.4', 'pylint==1.6.4', 'pyodbc==4.0.16', 'pyopenssl==17.0.0', 'pyparsing==2.1.4', 'pytest==3.0.7', 'python-dateutil==2.6.0', 'pytz==2017.2', 'pywavelets==0.5.2', 'pyyaml==3.12', 'pyzmq==16.0.2', 'qtawesome==0.4.4', 'qtconsole==4.3.0', 'qtpy==1.2.1', 'redis==2.10.5', 'requests==2.14.2', 'rope-py3k==0.9.4.post1', 'scikit-image==0.13.0', 'scikit-learn==0.18.1', 'scipy==0.19.0', 'seaborn==0.7.1', 'setuptools==27.2.0', 'simplegeneric==0.8.1', 'singledispatch==3.4.0.3', 'six==1.10.0', 'snowballstemmer==1.2.1', 'sockjs-tornado==1.0.3', 'sortedcollections==0.5.3', 'sortedcontainers==1.5.7', 'sphinx==1.5.6', 'spyder==3.1.4', 'sqlalchemy==1.1.9', 'statsmodels==0.8.0', 'sympy==1.0', 'tables==3.3.0', 'tblib==1.3.2', 'tensorflow==1.2.0rc1', 'terminado==0.6', 'testpath==0.3', 'tflearn==0.3.1', 'theano==0.9.0', 'toolz==0.8.2', 'tornado==4.5.1', 'traitlets==4.3.2', 'unicodecsv==0.14.1', 'wcwidth==0.1.7', 'werkzeug==0.12.2', 'wheel==0.29.0', 'widgetsnbextension==2.0.0', 'wrapt==1.10.10', 'xgboost==0.6', 'xlrd==1.0.0', 'xlsxwriter==0.9.6', 'xlwings==0.10.4', 'xlwt==1.2.0', 'zict==0.1.2']

I'll admit that I found a related solution, but it is not clear how it can be modified to fit my needs. The problem is that you can't do one hot on "first" column with values {a, b, c, ...} and then one-hot on "second" column that may also have values {a, b, c, ...} and label those columns by value. I would have a name clash. Pandas One hot encoding: Bundling together less frequent categories

like image 350
ldmtwo Avatar asked Dec 14 '22 22:12

ldmtwo


1 Answers

Consider the sample dataframe df

np.random.seed([3,1415])
df = pd.DataFrame(dict(
        name=np.random.choice(
            list('abcdefghij'), 1000,
            p=np.arange(10, 0, -1) / 55
        )
    ))
threshold = 60
counts = df.name.value_counts()
counts

a    197
b    166
c    139
d    119
f    107
e    105
g     72
h     53
i     27
j     15
Name: name, dtype: int64

Then replace and pd.get_dummies

repl = counts[counts <= threshold].index
print(pd.get_dummies(df.name.replace(repl, 'uncommon')))

     a  b  c  d  e  f  g  uncommon
0    0  0  1  0  0  0  0         0
1    0  0  1  0  0  0  0         0
2    0  0  1  0  0  0  0         0
3    0  0  1  0  0  0  0         0
4    0  0  1  0  0  0  0         0
5    1  0  0  0  0  0  0         0
6    0  0  0  0  0  0  1         0
7    0  0  0  0  0  1  0         0
8    0  0  0  0  0  1  0         0
9    0  0  0  0  0  1  0         0
10   0  0  0  0  0  0  0         1
11   0  0  0  0  0  0  1         0
12   0  0  0  0  0  0  1         0
13   0  0  0  0  0  0  0         1
14   0  0  0  0  1  0  0         0
15   1  0  0  0  0  0  0         0
16   1  0  0  0  0  0  0         0
17   0  1  0  0  0  0  0         0
like image 66
piRSquared Avatar answered Dec 28 '22 07:12

piRSquared