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?
On Wed, Feb 13, 2019 at 6:50 PM Paul Hobson firstname.lastname@example.org 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)
a b diff
0 NaN 15.0 NaN 1 86.0 72.0 14.0 2 87.0 2.0 85.0 3 86.0 NaN NaN
So what I think you want would be the following:
a = pandas.Series([85, 86, 87, 86], name='a') b = pandas.Series([15, 72, 2, 3], name='b') result = a.subtract(b.shift()).dropna() print(result) 1 71.0 2 15.0 3 84.0 dtype: float64
On Wed, Feb 13, 2019 at 2:51 PM C W email@example.com wrote:
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])
Example 2: worked If I use values() method, it's converted to a Numpy object. And it works!
array([71, 15, 84])
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?
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