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