By using apply df.groupby([‘col1’, ‘col2’])[“col3”, “col4”].apply(lambda x : x.astype(int).sum()) Out[1257]: col3 col4 col1 col2 a c 2 4 d 1 2 b d 1 2 e 2 4 If you want to agg df.groupby([‘col1’, ‘col2’]).agg({‘col3′:’sum’,’col4′:’sum’})
apply applies the function to each group (your Species). Your function returns 1, so you end up with 1 value for each of 3 groups. agg aggregates each column (feature) for each group, so you end up with one value per column per group. Do read the groupby docs, they’re quite helpful. There are also … Read more
consider the dataframe df df = pd.DataFrame(dict(A=list(‘aabb’), B=[1, 2, 3, 4], C=[0, 9, 0, 9])) groupby is the standard use aggregater df.groupby(‘A’).mean() maybe you want these values broadcast across the whole group and return something with the same index as what you started with. use transform df.groupby(‘A’).transform(‘mean’) df.set_index(‘A’).groupby(level=”A”).transform(‘mean’) agg is used when you have specific … 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
Use value_counts with normalize=True: df[‘gender’].value_counts(normalize=True) * 100 The result is a fraction in range (0, 1]. We multiply by 100 here in order to get the %.
Apply a lambda and call sample with param frac: In [2]: df = pd.DataFrame({‘a’: [1,2,3,4,5,6,7], ‘b’: [1,1,1,0,0,0,0]}) grouped = df.groupby(‘b’) grouped.apply(lambda x: x.sample(frac=0.3)) Out[2]: a b b 0 6 7 0 1 2 3 1
You can use groupby by dates of column Date_Time by dt.date: df = df.groupby([df[‘Date_Time’].dt.date]).mean() Sample: df = pd.DataFrame({‘Date_Time’: pd.date_range(’10/1/2001 10:00:00′, periods=3, freq=’10H’), ‘B’:[4,5,6]}) print (df) B Date_Time 0 4 2001-10-01 10:00:00 1 5 2001-10-01 20:00:00 2 6 2001-10-02 06:00:00 print (df[‘Date_Time’].dt.date) 0 2001-10-01 1 2001-10-01 2 2001-10-02 Name: Date_Time, dtype: object df = df.groupby([df[‘Date_Time’].dt.date])[‘B’].mean() print(df) … Read more
Your syntax is wrong. Here’s the correct way: df.drop_duplicates(subset=[‘bio’, ‘center’, ‘outcome’]) Or in this specific case, just simply: df.drop_duplicates() Both return the following: bio center outcome 0 1 one f 2 1 two f 3 4 three f Take a look at the df.drop_duplicates documentation for syntax details. subset should be a sequence of column … Read more
try this: In [110]: (df.groupby(‘Company Name’) …..: .agg({‘Organisation Name’:’count’, ‘Amount’: ‘sum’}) …..: .reset_index() …..: .rename(columns={‘Organisation Name’:’Organisation Count’}) …..: ) Out[110]: Company Name Amount Organisation Count 0 Vifor Pharma UK Ltd 4207.93 5 or if you don’t want to reset index: df.groupby(‘Company Name’)[‘Amount’].agg([‘sum’,’count’]) or df.groupby(‘Company Name’).agg({‘Amount’: [‘sum’,’count’]}) Demo: In [98]: df.groupby(‘Company Name’)[‘Amount’].agg([‘sum’,’count’]) Out[98]: sum count Company … Read more
You can also use tail with groupby to get the last n values of the group: df.sort_values(‘date’).groupby(‘id’).tail(1) id product date 2 220 6647 2014-10-16 8 901 4555 2014-11-01 5 826 3380 2015-05-19