def test(df): df = df.copy(deep=True) df[‘tt’] = np.nan return df If you pass the dataframe into a function and manipulate it and return the same dataframe, you are going to get the same dataframe in modified version. If you want to keep your old dataframe and create a new dataframe with your modifications then by … Read more
Using list comprehension is maybe more “pythonic”. filtered = [x for x in set(lst) if x < C]
The OpenSSL module comes from the pyOpenSSL library. You can install it with Pip using a command like: pip install pyOpenSSL If it fails due to missing dependencies, see the instructions on installing pyOpenSSL’s dependencies in this answer to “How to install OpenSSL for Python”. (Also, as with all pip installs, depending upon your environment, … Read more
If you have the latest version of geopandas (0.3.0 as of writing), and the if df is a GeoDataFrame, you can use the x and y attributes on the geometry column: df[‘lon’] = df.point_object.x df[‘lat’] = df.point_object.y In general, if you have a column of shapely objects, you can also use apply to do what … Read more
The extend() method appends to the existing array and returns None. In your case, you are creating an array — [4, 5, 6] — on the fly, extending it and then discarding it. The variable b ends up with the return value of None.
On Mac OS X using homebrew: brew install cairo brew install pango
A minimal and complete working example of how to use this answer (tested in Python 3.6.1) # run.py from paramiko import SSHClient ssh = SSHClient() ssh.load_system_host_keys() ssh.connect(‘…’) print(‘started…’) stdin, stdout, stderr = ssh.exec_command(‘python -m example’, get_pty=True) for line in iter(stdout.readline, “”): print(line, end=””) print(‘finished.’) and # example.py, at the server import time for x in … Read more
You can check that using to_numeric and coercing errors: pd.to_numeric(df[‘column’], errors=”coerce”).notnull().all() For all columns, you can iterate through columns or just use apply df.apply(lambda s: pd.to_numeric(s, errors=”coerce”).notnull().all()) E.g. df = pd.DataFrame({‘col’ : [1,2, 10, np.nan, ‘a’], ‘col2’: [‘a’, 10, 30, 40 ,50], ‘col3’: [1,2,3,4,5.0]}) Outputs col False col2 False col3 True dtype: bool
You can use Python’s *args syntax for this: >>> a = range(20) >>> b = (5, 12) >>> a[slice(*b)] [5, 6, 7, 8, 9, 10, 11] Basically, you’re telling Python to unpack the tuple b into individual elements and pass each of those elements to the slice() function as individual arguments.
import pandas as pd data = pd.DataFrame([ (‘Q1′,’Blue’,100), (‘Q1′,’Green’,300), (‘Q2′,’Blue’,200), (‘Q2′,’Green’,350), (‘Q3′,’Blue’,300), (‘Q3′,’Green’,400), (‘Q4′,’Blue’,400), (‘Q4′,’Green’,450), ], columns=[‘quarter’, ‘company’, ‘value’] ) data = data.set_index([‘quarter’, ‘company’]).value data.unstack().plot(kind=’bar’, stacked=True) If you don’t want to stack your bar chart: data.unstack().plot(kind=’bar’)