TypeError: 'DataFrameGroupBy' object does not support item assignment 这个错误信息表明你尝试对一个 DataFrameGroupBy 对象进行元素赋值操作,但这是不被支持的。以下是针对这个问题的详细解答: 错误信息的含义: 这个错误提示你正在尝试对一个 DataFrameGroupBy 对象进行元素赋值,但 Da
Perhaps, this is an expected behavior, but I can't seem to find a documentation for it. Expected Behavior I would expect a KeyError. The following is a pytest code that describes my expectations. The test passes import pandas as pd import pytest def test_dataframe_groupby(): df = pd.Da...
whenfuncacts as a transform, groupby.apply should produce a result that has the same index as the original. The result for the nth value with group key a in the input dataframe should be the nth value with group key a in the output dataframe. ...
DataFrame({'a':[1,1,1,2,2,2,3,3,3], 'b':[6,5,4,4,6,5,3,3,3]}) df['rk_min']= df.groupby(by=['a'])['b'].rank(ascending=True, method='min', na_option='bottom') df['rk_dense']= df.groupby(by=['a'])['b'].rank(ascending=True, method='dense', na_option...
Caused by the lines in DataFrame.reset_index if not inplace: return new_obj 👍 1 Member Author wesm commented Sep 19, 2016 This is technically an API change -- any user which has reset_index(inplace=True) on the last line of a IPython/Jupyter cell could hypothetically have GC ...
To acquire the meta from the input dataframe. The input data frame images_df has an index set already called image_id. I later do balanced_df = images_df.groupby(self.images_column_class_id) balanced_df = balanced_df.apply(self.balance_class, meta=balanced_schema) return balanced_df An...
xref #14139 for empty MI Hi everybody, in the current version renaming of MultiIndex DataFrames does not work. Lets take the following example: import datetime as DT import pandas as pd df = pd.DataFrame({ 'Branch' : 'A A A A A B'.split(...
The Anndata is named as adata. I added .obs object from a dataframe: adata.obs = adata.obs.merge(mutation, on='ID', how='left') And then got subgroups of adata by another .obs object: a = adata[adata.obs['tissue'] == 'skin'] View of AnnD...
f = pandas.DataFrame(data = [[0, 0]], columns = ['A', 'B']).astype('Int64') f['C'] = f['A'] / f['B'] this should return A B C 0 0 0 <NA> and not A B C 0 0 0 NaN phofl added Numeric Operations and removed Closing Candidate labels Feb 21, 2021 Member ...