python - Linear interpolation lookup of a dataframe - Stack Overflow

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I have a dataframe using pandas, something like:

d = {'X': [1, 2, 3], 'Y': [220, 187, 170]}
df = pd.DataFrame(data=d)

the dataframe ends up like

X	Y
1	220
2	187
3	170

I have a dataframe using pandas, something like:

d = {'X': [1, 2, 3], 'Y': [220, 187, 170]}
df = pd.DataFrame(data=d)

the dataframe ends up like

X	Y
1	220
2	187
3	170

I can get the y value for an x value of 1.0 using

df[df['X'] == 1.0]['Y']

which returns 220

But is there a way to get a linearly interpolated value of Y for an X value between values of X? For example, if I had an X value of 1.5, I would want it to return an interpolated value of 203.5.

I tried the interpolate function, but it permanently adjusts the data in the dataframe. I could also write a separate function that would calculate this, but I was wondering if there was a native function in pandas.

• python
• pandas

Share Improve this question Follow edited Nov 22, 2024 at 18:37 ouroboros1 13.6k77 gold badges3535 silver badges5353 bronze badges asked Nov 22, 2024 at 18:32 Graham TattersallGraham Tattersall 3311 silver badge22 bronze badges

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1 Answer 1

Sorted by: Reset to default 2

You can use np.interp for this:

import numpy as np

X_value = 1.5

np.interp(X_value, df['X'], df['Y'])
# 203.5

Make sure that df['X'] is monotonically increasing.

You can use the left and right parameters to customize the return value for out-of-bounds values:

np.interp(0.5, df['X'], df['Y'], left=np.inf, right=-np.inf)
# inf

# because 0.5 < df['X'].iloc[0]

By default, out-of-bounds values will correspond to the closest valid X value:

np.interp(10, df['X'], df['Y'])
# 170

# i.e., match for df['X'].iloc[-1]

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