It looks like a transposed matrix: df.values.T.tolist() [list(l) for l in zip(*df.values)] [[0, 0, 1, 1, 1, 0], [1, 1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 1], [0, 1, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 1, 1, 1, 0], [1, 1, 0, 0, 0, 1]]
I think you are intrested in distance_matrix. For example: Create data: import pandas as pd from scipy.spatial import distance_matrix data = [[5, 7], [7, 3], [8, 1]] ctys = [‘Boston’, ‘Phoenix’, ‘New York’] df = pd.DataFrame(data, columns=[‘xcord’, ‘ycord’], index=ctys) Output: xcord ycord Boston 5 7 Phoenix 7 3 New York 8 1 Using the distance … Read more
You can set inplace to True (default is False): area.replace(0, np.nan, inplace=True) See examples in docs.
This approach uses .groupby() and .ngroup() (new in Pandas 0.20.2) to create the id column: df[‘id’] = df.groupby([‘LastName’,’FirstName’]).ngroup() >>> df First Second id 0 Tom Jones 0 1 Tom Jones 0 2 David Smith 1 3 Alex Thompson 2 4 Alex Thompson 2 I checked timings and, for the small dataset in this example, Alexander’s … Read more
By using ngroups df.groupby([‘col_a’, ‘col_b’]).ngroups Out[101]: 6 Or using set len(set(zip(df[‘col_a’],df[‘col_b’]))) Out[106]: 6
You need groupby + size + Series.reset_index: df = df.groupby([‘Col1’, ‘Col2’]).size().reset_index(name=”Freq”) print (df) Col1 Col2 Freq 0 1 1 1 1 1 2 3 2 3 4 2
reindex+ repeat df.reindex(df.index.repeat(df.persons)) Out[951]: code . role ..1 persons 0 123 . Janitor . 3 0 123 . Janitor . 3 0 123 . Janitor . 3 1 123 . Analyst . 2 1 123 . Analyst . 2 2 321 . Vallet . 2 2 321 . Vallet . 2 3 321 . Auditor … Read more
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