How do numpy functions operate on pandas objects internally?
numpy, pandas, python
Solution
If you step through this:
--Call--
> d:\winpython-64bit-3.4.3.5\python-3.4.3.amd64\lib\site-packages\numpy\core\fromnumeric.py(2796)mean()
-> def mean(a, axis=None, dtype=None, out=None, keepdims=False):
(Pdb) s
> d:\winpython-64bit-3.4.3.5\python-3.4.3.amd64\lib\site-packages\numpy\core\fromnumeric.py(2877)mean()
-> if type(a) is not mu.ndarray:
(Pdb) s
> d:\winpython-64bit-3.4.3.5\python-3.4.3.amd64\lib\site-packages\numpy\core\fromnumeric.py(2878)mean()
-> try:
(Pdb) s
> d:\winpython-64bit-3.4.3.5\python-3.4.3.amd64\lib\site-packages\numpy\core\fromnumeric.py(2879)mean()
-> mean = a.mean
You can see that the `type` is not a `ndarray` so it tries to call `a.mean` which in this case would be `df.mean()`:
In [6]:
df.mean()
Out[6]:
0 0.572999
1 0.468268
dtype: float64
This is why the output is different
Code to reproduce above:
In [3]:
a = np.random.rand(4,2)
a
Out[3]:
array([[ 0.96750329, 0.67623187],
[ 0.44025179, 0.97312747],
[ 0.07330062, 0.18341157],
[ 0.81094166, 0.04030253]])
In [4]:
np.mean(a)
Out[4]:
0.52063384885403818
In [5]:
df = pd.DataFrame(data=a, index=range(np.shape(a)[0]),
columns=range(np.shape(a)[1]))
df
Out[5]:
0 1
0 0.967503 0.676232
1 0.440252 0.973127
2 0.073301 0.183412
3 0.810942 0.040303
numpy output:
In [7]:
np.mean(df)
Out[7]:
0 0.572999
1 0.468268
dtype: float64
If you'd called `.values` to return a `np` array then the output is the same:
In [8]:
np.mean(df.values)
Out[8]:
0.52063384885403818
Problem
Numpy functions, eg np.mean(), np.var(), etc, accept an array-like argument, like np.array, or list, etc. But passing in a pandas dataframe also works. This means that a pandas dataframe can indeed disguise itself as a numpy array, which I find a little strange (despite knowing the fact that the underlying values of a df are indeed numpy arrays). For an object to be an array-like, I thought that it should be slicable using integer indexing in the way a numpy array is sliced. So for instance df[1:3, 2:3] should work, but it would lead to an error. So, possibly a dataframe gets converted into a numpy array when it goes inside the function. But if that is the case then why does np.mean(numpy_array) lead to a different result than that of np.mean(df)? ``` a = np.random.rand(4,2) a Out[13]: array([[ 0.86688862, 0.09682919], [ 0.49629578, 0.78263523], [ 0.83552411, 0.71907931], [ 0.95039642, 0.71795655]]) np.mean(a) Out[14]: 0.68320065182041034 ``` gives a different result than what the below gives... ``` df = pd.DataFrame(data=a, index=range(np.shape(a)[0]), columns=range(np.shape(a)[1])) df Out[18]: 0 1 0 0.866889 0.096829 1 0.496296 0.782635 2 0.835524 0.719079 3 0.950396 0.717957 np.mean(df) Out[21]: 0 0.787276 1 0.579125 dtype: float64 ``` The former output is a single number, whereas the latter is a column-wise mean. How does a numpy function know about the make of a dataframe?