Elegant way to create empty pandas DataFrame with NaN of type float

Question

I want to create a Pandas DataFrame filled with NaNs. During my research I found an answer:

import pandas as pd

df = pd.DataFrame(index=range(0,4),columns=['A'])

This code results in a DataFrame filled with NaNs of type "object". So they cannot be used later on for example with the interpolate() method. Therefore, I created the DataFrame with this complicated code (inspired by this answer):

import pandas as pd
import numpy as np

dummyarray = np.empty((4,1))
dummyarray[:] = np.nan

df = pd.DataFrame(dummyarray)

This results in a DataFrame filled with NaN of type "float", so it can be used later on with interpolate(). Is there a more elegant way to create the same result?

Answer 1

Simply pass the desired representative as a scalar first argument, like 0, math.inf or, in this case, np.nan. The constructor then initializes the value array to the size specified by index and columns:

 >>> df = pd.DataFrame(np.nan, index=[0,1,2,3], columns=['A'])
 >>> df.dtypes
 A    float64
 dtype: object

Answer 2

You could specify the dtype directly when constructing the DataFrame:

>>> df = pd.DataFrame(index=range(0,4),columns=['A'], dtype='float')
>>> df.dtypes
A    float64
dtype: object

Specifying the dtype forces Pandas to try creating the DataFrame with that type, rather than trying to infer it.

Answer 3

Hope this can help!

 pd.DataFrame(np.nan, index = np.arange(<num_rows>), columns = ['A'])

Answer 4

You can try this line of code:

pdDataFrame = pd.DataFrame([np.nan] * 7)

This will create a pandas dataframe of size 7 with NaN of type float:

if you print pdDataFrame the output will be:

     0
0   NaN
1   NaN
2   NaN
3   NaN
4   NaN
5   NaN
6   NaN

Also the output for pdDataFrame.dtypes is:

0    float64
dtype: object

Answer 5

For multiple columns you can do:

df = pd.DataFrame(np.zeros([nrow, ncol])*np.nan)