Hi Paul,
Thanks for your response! I did not find a Pandas list for users, only for developers. I'd love to be on there.
result = a.subtract(b.shift()).dropna()This seems verbose, several layers of parenthesis follow by a dot method. I'm new to Python, I thought Python code would be pity and short. Is this what everyone will write?
Thank you!
_______________________________________________On Wed, Feb 13, 2019 at 6:50 PM Paul Hobson <pmhobson@gmail.com> wrote:
This is more a question for the pandas list, but since i'm here i'll take a crack.
- numpy aligns arrays by position.
- pandas aligns by label.
So what you did in pandas is roughly equivalent to the following:
a = pandas.Series([85, 86, 87, 86], name='a').iloc[1:4].to_frame()
b = pandas.Series([15, 72, 2, 3], name='b').iloc[0:3].to_frame()
result = a.join(b,how='outer').assign(diff=lambda df: df['a'] - df['b'])print(result)
_______________________________________________On Wed, Feb 13, 2019 at 2:51 PM C W <tmrsg11@gmail.com> wrote:
_______________________________________________Dear list,
I have the following to Pandas Series: a, b. I want to slice and then subtract. Like this: a[1:4] - b[0:3]. Why does it give me NaN? But it works in Numpy.
Example 1: did not work>>>a = pd.Series([85, 86, 87, 86])
>>>b = pd.Series([15, 72, 2, 3])>>> a[1:4]-b[0:3] 0 NaN 1 14.0 2 85.0 3 NaN
>>> type(a[1:4])<class 'pandas.core.series.Series'>
Example 2: workedIf I use values() method, it's converted to a Numpy object. And it works!>>> a.values[1:4]-b.values[0:3]array([71, 15, 84])>>> type(a.values[1:4])<class 'numpy.ndarray'>
What's the reason that Pandas in example 1 did not work? Isn't Numpy built on top of Pandas? So, why is everything ok in Numpy, but not in Pandas?
Thanks in advance!
NumPy-Discussion mailing list
NumPy-Discussion@python.org
https://mail.python.org/mailman/listinfo/numpy-discussion
SciPy-User mailing list
SciPy-User@python.org
https://mail.python.org/mailman/listinfo/scipy-user
SciPy-User mailing list
SciPy-User@python.org
https://mail.python.org/mailman/listinfo/scipy-user