The only way to do this would be to include C in your groupby (the groupby function can accept a list). Give this a try: df.groupby([‘A’,’C’])[‘B’].sum() One other thing to note, if you need to work with df after the aggregation you can also use the as_index=False option to return a dataframe object. This one … Read more
You can use the get_group method: In [21]: gb.get_group(‘foo’) Out[21]: A B C 0 foo 1.624345 5 2 foo -0.528172 11 4 foo 0.865408 14 Note: This doesn’t require creating an intermediary dictionary / copy of every subdataframe for every group, so will be much more memory-efficient than creating the naive dictionary with dict(iter(gb)). This … Read more
iloc df1 = datasX.iloc[:, :72] df2 = datasX.iloc[:, 72:] (iloc docs)
Try pd.isna(): In [7]: pd.isna(df.iloc[1,0]) Out[7]: True AKA pd.isnull
Why not just do this? df[‘NewCol’] = df.apply(lambda x: segmentMatch(x[‘TimeCol’], x[‘ResponseCol’]), axis=1) Rather than trying to pass the column as an argument as in your example, we now simply pass the appropriate entries in each row as argument, and store the result in ‘NewCol’.
As of pandas 2.0, append (previously deprecated) was removed. You need to use concat instead (for most applications): df = pd.concat([df, pd.DataFrame([new_row])], ignore_index=True) As noted by @cottontail, it’s also possible to use loc, although this only works if the new index is not already present in the DataFrame (typically, this will be the case if … Read more
You can use groupby’s size import pandas as pd # load the sample data data = {‘Group’: [‘Short’, ‘Short’, ‘Moderate’, ‘Moderate’, ‘Tall’], ‘Size’: [‘Small’, ‘Small’, ‘Medium’, ‘Small’, ‘Large’]} df = pd.DataFrame(data) Option 1: dfg = df.groupby(by=[“Group”, “Size”]).size() # which results in a pandas.core.series.Series Group Size Moderate Medium 1 Small 1 Short Small 2 Tall Large … Read more
Assuming your DataFrame is in df: df.Temp_Rating.fillna(df.Farheit, inplace=True) del df[‘Farheit’] df.columns=”File heat Observations”.split() First replace any NaN values with the corresponding value of df.Farheit. Delete the ‘Farheit’ column. Then rename the columns. Here’s the resulting DataFrame: File heat Observations 0 1 YesQ 75 1 1 NoR 115 2 1 YesA 63 3 1 NoT 41 … Read more
If you only have two choices to select from then use np.where: df[‘color’] = np.where(df[‘Set’]==’Z’, ‘green’, ‘red’) For example, import pandas as pd import numpy as np df = pd.DataFrame({‘Type’:list(‘ABBC’), ‘Set’:list(‘ZZXY’)}) df[‘color’] = np.where(df[‘Set’]==’Z’, ‘green’, ‘red’) print(df) yields Set Type color 0 Z A green 1 Z B green 2 X B red 3 Y … Read more
I believe DataFrame.fillna() will do this for you. Link to Docs for a dataframe and for a Series. Example: In [7]: df Out[7]: 0 1 0 NaN NaN 1 -0.494375 0.570994 2 NaN NaN 3 1.876360 -0.229738 4 NaN NaN In [8]: df.fillna(0) Out[8]: 0 1 0 0.000000 0.000000 1 -0.494375 0.570994 2 0.000000 0.000000 … Read more