Try doing this: week_grouped = df.groupby(‘week’) week_grouped.sum().reset_index().to_csv(‘week_grouped.csv’) That’ll write the entire dataframe to the file. If you only want those two columns then, week_grouped = df.groupby(‘week’) week_grouped.sum().reset_index()[[‘week’, ‘count’]].to_csv(‘week_grouped.csv’) Here’s a line by line explanation of the original code: # This creates a “groupby” object (not a dataframe object) # and you store it in the … Read more
For Pandas 0.17+, use sort_values: df.groupby(‘col1’).size().sort_values(ascending=False) For pre-0.17, you can use size().order(): df.groupby(‘col1’).size().order(ascending=False)
this code: data[‘2013-08-17′].boxplot(by=’SPECIES’) Will not work, as boxplot is a function for a DataFrame and not a Series. While in Pandas > 0.18.1 the boxplot function has the argument columns which defines from what column the data is taken from. So data.boxplot(column=’2013-08-17′,by=’SPECIES’) should return the desired result. An example with the Iris dataset: import pandas … Read more
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)
I think you should be able to use the nameattribute: temp_dataframe.groupby(level=0,axis=0).apply(lambda x: foo(x.name, x)) should work, example: In [132]: df = pd.DataFrame({‘a’:list(‘aabccc’), ‘b’:np.arange(6)}) df Out[132]: a b 0 a 0 1 a 1 2 b 2 3 c 3 4 c 4 5 c 5 In [134]: df.groupby(‘a’).apply(lambda x: print(‘name:’, x.name, ‘\nsubdf:’,x)) name: a subdf: … Read more
What pipe does is to allow you to pass a callable with the expectation that the object that called pipe is the object that gets passed to the callable. With apply we assume that the object that calls apply has subcomponents that will each get passed to the callable that was passed to apply. In … Read more
Try this: In [6]: df[‘date’] = pd.to_datetime(df[‘date’]) In [7]: df Out[7]: date Revenue 0 2017-06-02 100 1 2017-05-23 200 2 2017-05-20 300 3 2017-06-22 400 4 2017-06-21 500 In [59]: df.groupby(df[‘date’].dt.strftime(‘%B’))[‘Revenue’].sum().sort_values() Out[59]: date May 500 June 1000
I think it is Automatic exclusion of ‘nuisance’ columns, what described here. Sample: df = pd.DataFrame({‘C’: {0: -0.91985400000000006, 1: -0.042379, 2: 1.2476419999999999, 3: -0.00992, 4: 0.290213, 5: 0.49576700000000001, 6: 0.36294899999999997, 7: 1.548106}, ‘A’: {0: ‘foo’, 1: ‘bar’, 2: ‘foo’, 3: ‘bar’, 4: ‘foo’, 5: ‘bar’, 6: ‘foo’, 7: ‘foo’}, ‘B’: {0: ‘one’, 1: ‘one’, 2: … Read more
First groupby the key1 column: In [11]: g = df.groupby(‘key1′) and then for each group take the subDataFrame where key2 equals ‘one’ and sum the data1 column: In [12]: g.apply(lambda x: x[x[‘key2’] == ‘one’][‘data1′].sum()) Out[12]: key1 a 0.093391 b 1.468194 dtype: float64 To explain what’s going on let’s look at the ‘a’ group: In [21]: … Read more
Compute a mask with isna, then group and find the sum: df.drop(‘CLASS’, 1).isna().groupby(df.CLASS, sort=False).sum().reset_index() CLASS FEATURE1 FEATURE2 FEATURE3 0 X 1.0 1.0 2.0 1 B 0.0 0.0 0.0 Another option is to subtract the size from the count using rsub along the 0th axis for index aligned subtraction: df.groupby(‘CLASS’).count().rsub(df.groupby(‘CLASS’).size(), axis=0) Or, g = df.groupby(‘CLASS’) g.count().rsub(g.size(), … Read more