What is the best way to create a zero-filled pandas data frame of a given size?

I have used:

zero_data = np.zeros(shape=(len(data),len(feature_list)))
d = pd.DataFrame(zero_data, columns=feature_list)

Is there a better way to do it?

You can try this:

d = pd.DataFrame(0, index=np.arange(len(data)), columns=feature_list)

It’s best to do this with numpy in my opinion

import numpy as np
import pandas as pd
d = pd.DataFrame(np.zeros((N_rows, N_cols)))

If you would like the new data frame to have the same index and columns as an existing data frame, you can just multiply the existing data frame by zero:

df_zeros = df * 0

If the existing data frame contains NaNs or non-numeric values you can instead apply a function to each cell that will just return 0:

df_zeros = df.applymap(lambda x: 0)

Similar to @Shravan, but without the use of numpy:

  height = 10
  width = 20
  df_0 = pd.DataFrame(0, index=range(height), columns=range(width))

Then you can do whatever you want with it:

post_instantiation_fcn = lambda x: str(x)
df_ready_for_whatever = df_0.applymap(post_instantiation_fcn)

Assuming having a template DataFrame, which one would like to copy with zero values filled here…

If you have no NaNs in your data set, multiplying by zero can be significantly faster:

In [19]: columns = ["col{}".format(i) for i in xrange(3000)]                                                                                       

In [20]: indices = xrange(2000)

In [21]: orig_df = pd.DataFrame(42.0, index=indices, columns=columns)

In [22]: %timeit d = pd.DataFrame(np.zeros_like(orig_df), index=orig_df.index, columns=orig_df.columns)
100 loops, best of 3: 12.6 ms per loop

In [23]: %timeit d = orig_df * 0.0
100 loops, best of 3: 7.17 ms per loop

Improvement depends on DataFrame size, but never found it slower.

And just for the heck of it:

In [24]: %timeit d = orig_df * 0.0 + 1.0
100 loops, best of 3: 13.6 ms per loop

In [25]: %timeit d = pd.eval('orig_df * 0.0 + 1.0')
100 loops, best of 3: 8.36 ms per loop

But:

In [24]: %timeit d = orig_df.copy()
10 loops, best of 3: 24 ms per loop

EDIT!!!

Assuming you have a frame using float64, this will be the fastest by a huge margin! It is also able to generate any value by replacing 0.0 to the desired fill number.

In [23]: %timeit d = pd.eval('orig_df > 1.7976931348623157e+308 + 0.0')
100 loops, best of 3: 3.68 ms per loop

Depending on taste, one can externally define nan, and do a general solution, irrespective of the particular float type:

In [39]: nan = np.nan
In [40]: %timeit d = pd.eval('orig_df > nan + 0.0')
100 loops, best of 3: 4.39 ms per loop

If you already have a dataframe, this is the fastest way:

In [1]: columns = ["col{}".format(i) for i in range(10)]
In [2]: orig_df = pd.DataFrame(np.ones((10, 10)), columns=columns)
In [3]: %timeit d = pd.DataFrame(np.zeros_like(orig_df), index=orig_df.index, columns=orig_df.columns)
10000 loops, best of 3: 60.2 µs per loop

Compare to:

In [4]: %timeit d = pd.DataFrame(0, index = np.arange(10), columns=columns)
10000 loops, best of 3: 110 µs per loop

In [5]: temp = np.zeros((10, 10))
In [6]: %timeit d = pd.DataFrame(temp, columns=columns)
10000 loops, best of 3: 95.7 µs per loop