Most efficient way to forward-fill NaN values in numpy array

Question

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 1

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.]])

Solution 2

Update: As pointed out by financial_physician in the comments, my initially proposed solution can simply be exchanged with ffill on the reversed array and then reversing the result. There is no relevant performance loss. My initial solution seems to be 2% or 3% faster according to %timeit. I updated the code example below but left my initial text as it was.

---

For those that came here looking for the backward-fill of NaN values, I modified the solution provided by Divakar above to do exactly that. The trick is that you have to do the accumulation on the reversed array using the minimum except for the maximum.

Here is the code:

# ffill along axis 1, as provided in the answer by Divakar
def ffill(arr):
    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

# Simple solution for bfill provided by financial_physician in comment below
def bfill(arr): 
    return ffill(arr[:, ::-1])[:, ::-1]

# My outdated modification of Divakar's answer to do a backward-fill
def bfill_old(arr):
    mask = np.isnan(arr)
    idx = np.where(~mask, np.arange(mask.shape[1]), mask.shape[1] - 1)
    idx = np.minimum.accumulate(idx[:, ::-1], axis=1)[:, ::-1]
    out = arr[np.arange(idx.shape[0])[:,None], idx]
    return out


# Test both functions
arr = np.array([[5, np.nan, np.nan, 7, 2],
                [3, np.nan, 1, 8, np.nan],
                [4, 9, 6, np.nan, np.nan]])
print('Array:')
print(arr)

print('\nffill')
print(ffill(arr))

print('\nbfill')
print(bfill(arr))

Output:

Array:
[[ 5. nan nan  7.  2.]
 [ 3. nan  1.  8. nan]
 [ 4.  9.  6. nan nan]]

ffill
[[5. 5. 5. 7. 2.]
 [3. 3. 1. 8. 8.]
 [4. 9. 6. 6. 6.]]

bfill
[[ 5.  7.  7.  7.  2.]
 [ 3.  1.  1.  8. nan]
 [ 4.  9.  6. nan nan]]

Edit: Update according to comment of MS_

Solution 3

Use Numba. This should give a significant speedup:

import numba
@numba.jit
def loops_fill(arr):
    ...

Solution 4

I liked Divakar's answer on pure numpy. Here's a generalized function for n-dimensional arrays:

def np_ffill(arr, axis):
    idx_shape = tuple([slice(None)] + [np.newaxis] * (len(arr.shape) - axis - 1))
    idx = np.where(~np.isnan(arr), np.arange(arr.shape[axis])[idx_shape], 0)
    np.maximum.accumulate(idx, axis=axis, out=idx)
    slc = [np.arange(k)[tuple([slice(None) if dim==i else np.newaxis
        for dim in range(len(arr.shape))])]
        for i, k in enumerate(arr.shape)]
    slc[axis] = idx
    return arr[tuple(slc)]

AFIK pandas can only work with two dimensions, despite having multi-index to make up for it. The only way to accomplish this would be to flatten a DataFrame, unstack desired level, restack, and finally reshape as original. This unstacking/restacking/reshaping, with the pandas sorting involved, is just unnecessary overhead to achieve the same result.

Testing:

def random_array(shape):
    choices = [1, 2, 3, 4, np.nan]
    out = np.random.choice(choices, size=shape)
    return out

ra = random_array((2, 4, 8))
print('arr')
print(ra)
print('\nffull')
print(np_ffill(ra, 1))
raise SystemExit

Output:

arr
[[[ 3. nan  4.  1.  4.  2.  2.  3.]
  [ 2. nan  1.  3. nan  4.  4.  3.]
  [ 3.  2. nan  4. nan nan  3.  4.]
  [ 2.  2.  2. nan  1.  1. nan  2.]]

 [[ 2.  3.  2. nan  3.  3.  3.  3.]
  [ 3.  3.  1.  4.  1.  4.  1. nan]
  [ 4.  2. nan  4.  4.  3. nan  4.]
  [ 2.  4.  2.  1.  4.  1.  3. nan]]]

ffull
[[[ 3. nan  4.  1.  4.  2.  2.  3.]
  [ 2. nan  1.  3.  4.  4.  4.  3.]
  [ 3.  2.  1.  4.  4.  4.  3.  4.]
  [ 2.  2.  2.  4.  1.  1.  3.  2.]]

 [[ 2.  3.  2. nan  3.  3.  3.  3.]
  [ 3.  3.  1.  4.  1.  4.  1.  3.]
  [ 4.  2.  1.  4.  4.  3.  1.  4.]
  [ 2.  4.  2.  1.  4.  1.  3.  4.]]]

Solution 5

I like Divakar's answer, but it doesn't work for an edge case where a row starts with np.nan, like the arr below

arr = np.array([[9, np.nan, 4, np.nan, 6, 6, 7, 2, 3, np.nan],
[ np.nan, 5, 5, 6, 5, 3, 2, 1, np.nan, 10]])

The output using Divakar's code would be:

[[ 9.  9.  4.  4.  6.  6.  7.  2.  3.  3.]
 [nan  4.  5.  6.  5.  3.  2.  1.  1. 10.]]

Divakar's code can be simplified a bit, and the simplified version solves this issue at the same time:

arr[np.isnan(arr)] = arr[np.nonzero(np.isnan(arr))[0], np.nonzero(np.isnan(arr))[1]-1]

In case of several np.nans in a row (either in the beginning or in the middle), just repeat this operation several times. For instance, if the array has 5 consecutive np.nans, the following code will "forward fill" all of them with the number before these np.nans:

for i in range(0, 5):
   value[np.isnan(value)] = value[np.nonzero(np.isnan(value))[0], np.nonzero(np.isnan(value))[1]-1]

View more solutions

Comments

[Xukrao]

A vectorized numpy-only solution, nice. Thanks! This solution indeed appears to be faster than the loop-based and pandas-based solutions (see timings in updated question).

[Divakar]

@Xukrao Yeah I just saw those, thanks for adding in those timing results! Good to see some speedups there!

[Xukrao]

Would Numba only speed up the loops-based solution? Or would it speed up the other solutions as well?

[shx2]

It is good for loops. It would not speed up functions implemented in numpy/pandas.

[Xukrao]

Thanks! I've included this solution in the timing comparison (see updated question). It looks like the addition of the numba decorator to the loop-based solution reduces its runtime by one order of magnitude.

[Xukrao]

I'm not sure I understand the purpose of this code. What exactly do you mean by 'problem of having leading np.nan after forward-filling'?

[christian_bock]

In the example array in the beginning of the threat, each entry begins with a non nan. Some people might find themselves dealing with a data set that requires backward filling because forward filling will leave the first entries untouched. So I thought it might be useful to present a solution in this threat.

[MS_]

idx = np.where(~mask, np.arange(mask.shape[1]), mask.shape[0] + 1) in bfill should be idx = np.where(~mask, np.arange(mask.shape[1]), mask.shape[1] - 1)

[user189035]

How do you adapt this solution to the case arr is a one dimensional numpy array? Like numpy.array([0.83, 0.83, 0.83, 0.83, nan, nan, nan])?

[C8H10N4O2]

@user189035 replace mask.shape[1] with mask.size and remove axis=1 and replace the last line with out = arr[idx]

[financial_physician]

I had a case where I built a second matrix for what I wanted to forward fill with. On the last line I just replaced arr with fillMatrix. My case was reducing resolution on time-series data, so I forward filled with the most recent entry

[LearnToGrow]

@Xukrao this does not work for a lot of cases. See my answer, I put an example where it fails!

[financial_physician]

Isn't flipping O(n) and you're doing it twice so wouldn't flipping, using forward fill, and then unflipping, be just as fast as your bfill method with the original array?

[cchwala]

Thanks! This is indeed a very good point. I did time your solution and mine using %%timeit and there is only a negligible but consistent difference, 10.3 µs (your solution) vs 9.95 µs (my solution). I will update my response accordingly.