Reputation: 21
I have a large data set, and I have some missing value, I want to fill the NAN values by the mean of the column before and after , and in certain cases i have NaN values consecutive in these case I want to replace all this nan values by the first value of non nan can found for examples : I should use a loop
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
19.0 NaN NaN NaN 29.0 30.0 NaN 16.0 15.0 16.0 17.0 NaN 28.0 30.0 NaN 28.0 18.0
The goal is for the data to look like this:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
19.0 29.0 29.0 29.0 29.0 30.0 23.0 16.0 15.0 16.0 17.0 22.5 28.0 30.0 29 28.0 18.0
Upvotes: 2
Views: 1094
Reputation: 150735
Let's try:
# where df is not null
s = df.notna()
# check for `NaN` with valid left and right:
mask = s.shift(1, axis=1) & s.shift(-1, axis=1)
# fill as required
df[:] = np.where(mask, df.interpolate(axis=1), df.bfill(axis=1).ffill(axis=1))
Output:
0 1 2 3 4 5 6 7 8 9 10 11 \
0 19.0 29.0 29.0 29.0 29.0 30.0 23.0 16.0 15.0 16.0 17.0 22.5
12 13 14 15 16
0 28.0 30.0 29.0 28.0 18.0
Upvotes: 3
Reputation: 1224
Let
import numpy as np
import pandas as pd
a = "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 19.0 NaN NaN NaN 29.0 30.0 NaN 16.0 15.0 16.0 17.0 NaN 28.0 30.0 NaN 28.0 18.0"
l = np.array([int(float(e)) if e != 'NaN' else np.nan for e in a.split(' ')])
Then what you are looking for could be accomplished with
subset_ranges = [0, 3]
replacements = {}
for i in range(len(l)-1):
subset = l[subset_ranges[0]: subset_ranges[1]]
if pd.isnull(subset[1]) and not pd.isnull(subset[0]) and not pd.isnull(subset[2]):
replacements[subset_ranges[0]+1] = np.nanmean(subset)
subset_ranges[0] += 1
subset_ranges[1] += 1
l = np.array([e if i not in replacements.keys() else replacements[i] for i, e in enumerate(l)])
df = pd.DataFrame(l.reshape(-1, 1))
df.fillna(method='bfill', inplace=True)
Upvotes: 0