Table of Contents

## Issue

## Example Problem

As a simple example, consider the numpy array `arr`

as defined below:

```
import numpy as np
arr = np.array([[5, np.nan, np.nan, 7, 2],
[3, np.nan, 1, 8, np.nan],
[4, 9, 6, np.nan, np.nan]])
```

where `arr`

looks like this in console output:

```
array([[ 5., nan, nan, 7., 2.],
[ 3., nan, 1., 8., nan],
[ 4., 9., 6., nan, nan]])
```

I would now like to row-wise ‘forward-fill’ the `nan`

values in array `arr`

. By that I mean replacing each `nan`

value with the nearest valid value from the left. The desired result would look like this:

```
array([[ 5., 5., 5., 7., 2.],
[ 3., 3., 1., 8., 8.],
[ 4., 9., 6., 6., 6.]])
```

## Tried thus far

I’ve tried using for-loops:

```
for row_idx in range(arr.shape[0]):
for col_idx in range(arr.shape[1]):
if np.isnan(arr[row_idx][col_idx]):
arr[row_idx][col_idx] = arr[row_idx][col_idx - 1]
```

I’ve also tried using a pandas dataframe as an intermediate step (since pandas dataframes have a very neat built-in method for forward-filling):

```
import pandas as pd
df = pd.DataFrame(arr)
df.fillna(method='ffill', axis=1, inplace=True)
arr = df.as_matrix()
```

Both of the above strategies produce the desired result, but I keep on wondering: wouldn’t a strategy that uses only numpy vectorized operations be the most efficient one?

## Summary

Is there another more efficient way to ‘forward-fill’ `nan`

values in numpy arrays? (e.g. by using numpy vectorized operations)

# Update: Solutions Comparison

I’ve tried to time all solutions thus far. This was my setup script:

```
import numba as nb
import numpy as np
import pandas as pd
def random_array():
choices = [1, 2, 3, 4, 5, 6, 7, 8, 9, np.nan]
out = np.random.choice(choices, size=(1000, 10))
return out
def loops_fill(arr):
out = arr.copy()
for row_idx in range(out.shape[0]):
for col_idx in range(1, out.shape[1]):
if np.isnan(out[row_idx, col_idx]):
out[row_idx, col_idx] = out[row_idx, col_idx - 1]
return out
@nb.jit
def numba_loops_fill(arr):
'''Numba decorator solution provided by shx2.'''
out = arr.copy()
for row_idx in range(out.shape[0]):
for col_idx in range(1, out.shape[1]):
if np.isnan(out[row_idx, col_idx]):
out[row_idx, col_idx] = out[row_idx, col_idx - 1]
return out
def pandas_fill(arr):
df = pd.DataFrame(arr)
df.fillna(method='ffill', axis=1, inplace=True)
out = df.as_matrix()
return out
def numpy_fill(arr):
'''Solution provided by Divakar.'''
mask = np.isnan(arr)
idx = np.where(~mask,np.arange(mask.shape[1]),0)
np.maximum.accumulate(idx,axis=1, out=idx)
out = arr[np.arange(idx.shape[0])[:,None], idx]
return out
```

followed by this console input:

```
%timeit -n 1000 loops_fill(random_array())
%timeit -n 1000 numba_loops_fill(random_array())
%timeit -n 1000 pandas_fill(random_array())
%timeit -n 1000 numpy_fill(random_array())
```

resulting in this console output:

```
1000 loops, best of 3: 9.64 ms per loop
1000 loops, best of 3: 377 µs per loop
1000 loops, best of 3: 455 µs per loop
1000 loops, best of 3: 351 µs per loop
```

## Solution

Here’s one approach –

```
mask = np.isnan(arr)
idx = np.where(~mask,np.arange(mask.shape[1]),0)
np.maximum.accumulate(idx,axis=1, out=idx)
out = arr[np.arange(idx.shape[0])[:,None], idx]
```

If you don’t want to create another array and just fill the NaNs in `arr`

itself, replace the last step with this –

```
arr[mask] = arr[np.nonzero(mask)[0], idx[mask]]
```

Sample input, output –

```
In [179]: arr
Out[179]:
array([[ 5., nan, nan, 7., 2., 6., 5.],
[ 3., nan, 1., 8., nan, 5., nan],
[ 4., 9., 6., nan, nan, nan, 7.]])
In [180]: out
Out[180]:
array([[ 5., 5., 5., 7., 2., 6., 5.],
[ 3., 3., 1., 8., 8., 5., 5.],
[ 4., 9., 6., 6., 6., 6., 7.]])
```

Answered By – Divakar

Answer Checked By – Mary Flores (BugsFixing Volunteer)