I believe this does what you are after: df.groupby(lambda x:x, axis=1).sum() Alternatively, between 3% and 15% faster depending on the length of the df: df.groupby(df.columns, axis=1).sum() EDIT: To extend this beyond sums, use .agg() (short for .aggregate()): df.groupby(df.columns, axis=1).agg(numpy.max)
df[‘purchase’] = 0 df.loc[df[‘dollars_spent’] > 0, ‘purchase’] = 1 or df[‘purchase’] = df[‘dollars_spent’].apply(lambda x: 1 if x > 0 else 0)
You can use isnull with mean for threshold and then remove columns by boolean indexing with loc (because remove columns), also need invert condition – so <.8 means remove all columns >=0.8: df = df.loc[:, df.isnull().mean() < .8] Sample: np.random.seed(100) df = pd.DataFrame(np.random.random((100,5)), columns=list(‘ABCDE’)) df.loc[:80, ‘A’] = np.nan df.loc[:5, ‘C’] = np.nan df.loc[20:, ‘D’] = … Read more
Use np.minimum: In [341]: df[‘MinNote’] = np.minimum(1,df[‘note’]) df Out[341]: session note minValue MinNote 0 1 0.726841 0.726841 0.726841 1 2 3.163402 3.163402 1.000000 2 3 2.844161 2.844161 1.000000 3 4 NaN NaN NaN Also min doesn’t understand array-like comparisons hence your error
def value_to_float(x): if type(x) == float or type(x) == int: return x if ‘K’ in x: if len(x) > 1: return float(x.replace(‘K’, ”)) * 1000 return 1000.0 if ‘M’ in x: if len(x) > 1: return float(x.replace(‘M’, ”)) * 1000000 return 1000000.0 if ‘B’ in x: return float(x.replace(‘B’, ”)) * 1000000000 return 0.0 df[‘col’] = … Read more
To get the highest values of a column you can use nlargest() : df[‘High’].nlargest(2) The above will give you the 2 highest values of column High. You can also use nsmallest() to get the lowest values.
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