As stated in another answer, the as_matrix method is deprecated since 0.23.0, so you should use to_numpy instead. However, I want to highlight the fact that as_matrix and to_numpy have different signatures: as_matrix takes a list of column names as one of its parameter, in case you want to limit the conversion to a subset … Read more
Installing matplotlib before installing pandas again made it work.
Taking elements from the solutions of @PallavBakshi and @Mike, the following works in Pandas >= 0.19: a.loc[a.index[-1], ‘a’] = 4.0 Just using iloc[-1, ‘a’] won’t work as ‘a’ is not a location.
If you’re trying to do this in ipython notebook, you can do: from IPython.display import display, HTML def pretty_print(df): return display( HTML( df.to_html().replace(“\\n”,”<br>”) ) )
The default aggfunc in pivot_table is np.sum and it doesn’t know what to do with strings and you haven’t indicated what the index should be properly. Trying something like: pivot_table = unified_df.pivot_table(index=[‘id’, ‘contact_id’], columns=”question”, values=”response_answer”, aggfunc=lambda x: ‘ ‘.join(x)) This explicitly sets one row per id, contact_id pair and pivots the set of response_answer values … Read more
use – import ssl ssl._create_default_https_context = ssl._create_unverified_context
You can do this: df[df[“col”].str.len() != 0] Example: import pandas as pd df = pd.DataFrame({“col”: [[1], [2, 3], [], [4, 5, 6], []]}, dtype=object) print(df[df[“col”].str.len() != 0]) # col # 0 [1] # 1 [2, 3] # 3 [4, 5, 6]
You can use .apply and datetime.replace, eg: import pandas as pd from datetime import datetime ps = pd.Series([datetime(2014, 1, 7), datetime(2014, 3, 13), datetime(2014, 6, 12)]) new = ps.apply(lambda dt: dt.replace(day=1)) Gives: 0 2014-01-01 1 2014-03-01 2 2014-06-01 dtype: datetime64[ns]
Just put ; after the code. It works only in Jupyter Notebook. plt.hist(…);
Using Pandas str.replace and regex: df[“new_column”] = df[‘review’].str.replace(‘[^\w\s]’,”)