The error comes up when you are trying to assign a list of numpy array of different length to a data frame, and it can be reproduced as follows: A data frame of four rows: Now trying to assign a list/array of two elements to it: Both errors out: ValueError: Length of values does not … Read more
For a dataframe df, one can use any of the following: len(df.index) df.shape[0] df[df.columns[0]].count() (== number of non-NaN values in first column) Code to reproduce the plot:
Use GroupBy.sum:
DataFrame.iterrows is a generator which yields both the index and row (as a Series):
Rather than create 2 temporary dfs you can just pass these as params within a dict using the DataFrame constructor: There are lots of ways to construct a df, see the docs
The pd.DataFrame constructor does not accept a dictionary view as data. You can convert to list instead. Here’s a minimal example: The docs do suggest this: data : numpy ndarray (structured or homogeneous), dict, or DataFrame Equivalently, you can use pd.DataFrame.from_dict, which accepts a dictionary directly:
df.to_numpy() is better than df.values, here’s why.* It’s time to deprecate your usage of values and as_matrix(). pandas v0.24.0 introduced two new methods for obtaining NumPy arrays from pandas objects: to_numpy(), which is defined on Index, Series, and DataFrame objects, and array, which is defined on Index and Series objects only. If you visit the v0.24 docs for .values, you will see a big red warning that says: Warning: We recommend … Read more
Edited for Pandas 0.22+ considering the deprecation of the use of dictionaries in a group by aggregation. We set up a very similar dictionary where we use the keys of the dictionary to specify our functions and the dictionary itself to rename the columns. option 1use agg ← link to docs option 2more for lessuse describe ← link to … Read more
For a dataframe df, one can use any of the following: len(df.index) df.shape[0] df[df.columns[0]].count() (== number of non-NaN values in first column) Code to reproduce the plot:
Use GroupBy.sum: