df = pd.DataFrame([[‘Jhon’,15,’A’],[‘Anna’,19,’B’],[‘Paul’,25,’D’]]) df. columns = [‘Name’,’Age’,’Grade’] df Out[472]: Name Age Grade 0 Jhon 15 A 1 Anna 19 B 2 Paul 25 D You can get the index of your row: i = df[((df.Name == ‘jhon’) &( df.Age == 15) & (df.Grade == ‘A’))].index and then drop it: df.drop(i) Out[474]: Name Age Grade 1 … Read more
In version 0.19.0 you can use parameter dtype=”category” in read_csv: data=”col1,col2,col3\na,b,1\na,b,2\nc,d,3″ df = pd.read_csv(pd.compat.StringIO(data), dtype=”category”) print (df) col1 col2 col3 0 a b 1 1 a b 2 2 c d 3 print (df.dtypes) col1 category col2 category col3 category dtype: object If want specify column for category use dtype with dictionary: df = pd.read_csv(pd.compat.StringIO(data), … Read more
Use dropna with parameter subset for specify column for check NaNs: data = data.dropna(subset=[‘sms’]) print (data) id city department sms category 1 2 lhr revenue good 1 Another solution with boolean indexing and notnull: data = data[data[‘sms’].notnull()] print (data) id city department sms category 1 2 lhr revenue good 1 Alternative with query: print (data.query(“sms … Read more
I tried everything you mentioned in a new environment using conda and I had another issue related to the version of ipywidgets (a bug found in Github with comments saying that got solved after using last version). I solved the problem I had installing last version of ipywidgets. Here is my process: Create a new … Read more
Use merge: print (pd.merge(df1, df2, left_on=’id’, right_on=’id1′, how=’left’).drop(‘id1’, axis=1)) id name count price rating 0 1 a 10 100.0 1.0 1 2 b 20 200.0 2.0 2 3 c 30 300.0 3.0 3 4 d 40 NaN NaN 4 5 e 50 500.0 5.0 Another solution is simple rename column: print (pd.merge(df1, df2.rename(columns={‘id1′:’id’}), on=’id’, how=’left’)) … Read more
You can transpose the single-row dataframe (which still results in a dataframe) and then squeeze the results into a series (the inverse of to_frame). df = pd.DataFrame([list(range(5))], columns=[“a{}”.format(i) for i in range(5)]) >>> df.squeeze(axis=0) a0 0 a1 1 a2 2 a3 3 a4 4 Name: 0, dtype: int64 Note: To accommodate the point raised by … Read more
The rough equivalent is .reset_index(), but it may not be helpful to think of it as the “opposite” of groupby(). You are splitting a string in to pieces, and maintaining each piece’s association with ‘family’. This old answer of mine does the job. Just set ‘family’ as the index column first, refer to the link … Read more
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’})
This functionality is not built into seaborn.countplot as far as I know – the order parameter only accepts a list of strings for the categories, and leaves the ordering logic to the user. This is not hard to do with value_counts() provided you have a DataFrame though. For example, import pandas as pd import seaborn … Read more
As mentioned in unutbu’s comment, groupby’s filter is the equivalent of SQL’S HAVING: In [11]: df = pd.DataFrame([[1, 2], [1, 3], [5, 6]], columns=[‘A’, ‘B’]) In [12]: df Out[12]: A B 0 1 2 1 1 3 2 5 6 In [13]: g = df.groupby(‘A’) # GROUP BY A In [14]: g.filter(lambda x: len(x) > … Read more