WebMar 5, 2013 · df.groupby ( ['client_id', 'date']).agg (pd.Series.mode) returns ValueError: Function does not reduce, since the first group returns a list of two (since there are two modes). (As documented here, if the first group returned a single mode this would work!) Two possible solutions for this case are: WebJun 30, 2016 · If you want to save even more ink, you don't need to use .apply () since .agg () can take a function to apply to each group: df.groupby ('id') ['words'].agg (','.join) OR # this way you can add multiple columns and different aggregates as needed. df.groupby ('id').agg ( {'words': ','.join}) Share Improve this answer Follow
Python 使用groupby和aggregate在第一个数据行的顶部创建一个 …
WebDataFrameGroupBy.agg(arg, *args, **kwargs) [source] ¶ Aggregate using callable, string, dict, or list of string/callables See also pandas.DataFrame.groupby.apply, pandas.DataFrame.groupby.transform, pandas.DataFrame.aggregate Notes WebMar 23, 2024 · You can drop the reset_index and then unstack. This will result in a Dataframe has the different counts for the different etnicities as columns. 1 minus the % of white employees will then yield the desired formula. df_agg = df_ethnicities.groupby ( ["Company", "Ethnicity"]).agg ( {"Count": sum}).unstack () percentatges = 1-df_agg [ … sidesync ita
pandas.core.groupby.DataFrameGroupBy.agg — pandas 2.0.0 …
WebWe can groupby the 'name' and 'month' columns, then call agg() functions of Panda’s DataFrame objects. The aggregation functionality provided by the agg() function allows … WebDec 14, 2024 · If your Pandas version is older than 0.25 then running the above code will give you the following error: TypeError: aggregate () missing 1 required positional argument: 'arg'. Now to do the aggregation for both value1 and value2, you will run this code: df_agg = df.groupby ( ['key1','key2'],as_index=False).agg ( {'value1': ['mean','count ... Web3 Answers. No need for the intermediate step. You can get a series with the string lengths like this: Now juut groupby key, and return the value indexed where the length of the string is largest using idxmax () In [33]: df.groupby ('key').agg (lambda x: x.loc [x.str.len ().idxmax ()]) Out [33]: text key 1 aaa 2 bbb 3 cc. the plot of black panther