Python: get the element-wise mean of multiple arra

Question

I have a 16x10 panda dataframe with 1x35000 arrays (or NaN) in each cell. I want to take the element-wise mean over rows for each column.

      1       2       3       ...       10
1    1x35000 1x35000 1x35000           1x35000

2    1x35000 NaN     1x35000           1x35000

3    1x35000 NaN     1x35000           NaN

...

16   1x35000 1x35000 NaN               1x35000

To avoid misunderstandings: take the first element of each array in the first column and take the mean. Then take the second element of each array in the first column and take the mean again. In the end I want to have a 1x10 dataframe with one 1x35000 array each per column. The array should be the element-wise mean of my former arrays.

      1       2       3       ...       10
1    1x35000 1x35000 1x35000           1x35000

Do you have an idea to get there elegantly preferably without for-loops?

Answer 1

Approach #1 : Loopy

Given the mixed dtype input data, we might want to loop through for performance efficiency. So, looping with explicit loops or under-the-hood usages of .apply/.applymap would be the solutions that could be suggested.

Here's one way looping through columns -

mask = ~df.isnull().values
n = df.shape[1]
out = np.empty((1,n),dtype=object)
for i in range(n):
    out[0,i] = df.iloc[mask[:,i],i].mean()
df_out = pd.DataFrame(out)

Sample input, output -

In [326]: df
Out[326]: 
              0             1             2
0  [4, 0, 1, 6]  [4, 2, 2, 2]  [5, 3, 5, 4]
1           NaN  [0, 5, 6, 8]           NaN
2           NaN           NaN           NaN
3           NaN           NaN           NaN

In [327]: df_out
Out[327]: 
                      0                     1                     2
0  [4.0, 0.0, 1.0, 6.0]  [2.0, 3.5, 4.0, 5.0]  [5.0, 3.0, 5.0, 4.0]

---

Approach #2 : Vectorized

If you have to vectorize, here's one way using matrix-multiplication to replace the mean-reductions and that could bring about improvements for large data -

mask = ~df.isnull().values
v = np.vstack(df.values[mask])
r,c = np.where(mask)
n = df.shape[1]
pos_mask = c == np.arange(n)[:,None]
out = pos_mask.dot(v)/np.bincount(c).astype(float)[:,None]
df_out1 = pd.DataFrame(out)

Sample output -

In [328]: df_out1
Out[328]: 
     0    1    2    3
0  4.0  0.0  1.0  6.0
1  2.0  3.5  4.0  5.0
2  5.0  3.0  5.0  4.0

Approach #3 : Vectorized one more

Making use of np.add.reduceat to get those mean-reductions -

mask = ~df.T.isnull().values
v = np.vstack(df.values.T[mask])
count = mask.sum(1)
out0 = np.add.reduceat(v, np.r_[0,count.cumsum()[:-1]])
out = out0/count[:,None].astype(float)
df_out2 = pd.DataFrame(out)

Answer 2

Setup

np.random.seed([3,14159])
df = pd.DataFrame(
    np.random.randint(10, size=(3, 3, 5)).tolist(),
    list('XYZ'), list('ABC')
).applymap(np.array)

df.loc['X', 'B'] = np.nan
df.loc['Z', 'A'] = np.nan

df

---

                 A                B                C
X  [4, 8, 1, 1, 9]              NaN  [8, 2, 8, 4, 9]
Y  [4, 3, 4, 1, 5]  [1, 2, 6, 2, 7]  [7, 1, 1, 7, 8]
Z              NaN  [9, 3, 8, 7, 7]  [2, 6, 3, 1, 9]

---

Solution

g = df.stack().groupby(level=1)
g.apply(np.sum, axis=0) / g.size()

A                        [4.0, 5.5, 2.5, 1.0, 7.0]
B                        [5.0, 2.5, 7.0, 4.5, 7.0]
C    [5.66666666667, 3.0, 4.0, 4.0, 8.66666666667]
dtype: object

If you insist on the shape you presented

g = df.stack().groupby(level=1)
(g.apply(np.sum, axis=0) / g.size()).to_frame().T

                           A                          B                                              C
0  [4.0, 5.5, 2.5, 1.0, 7.0]  [5.0, 2.5, 7.0, 4.5, 7.0]  [5.66666666667, 3.0, 4.0, 4.0, 8.66666666667]