I think you could also just opt for DataFrame.replace.

df.replace(' ', '_', regex=True)

Outputs

      Person_1    Person_2     Person_3
0   John_Smith  Jane_Smith   Mark_Smith
1  Harry_Jones  Mary_Jones  Susan_Jones

From some rough benchmarking, it predictably seems like piRSquared's NumPy solution is indeed the fastest, for this small sample at least, followed by DataFrame.replace.

%timeit df.values[:] = np.core.defchararray.replace(df.values.astype(str), ' ', '_')
10000 loops, best of 3: 78.4 µs per loop

%timeit df.replace(' ', '_', regex=True)
1000 loops, best of 3: 932 µs per loop

%timeit df.stack().str.replace(' ', '_').unstack()
100 loops, best of 3: 2.29 ms per loop

Interestingly however, it appears that piRSquared's Pandas solution scales much better with larger DataFrames than DataFrame.replace, and even outperforms the NumPy solution.

>>> df = pd.DataFrame([['John Smith', 'Jane Smith', 'Mark Smith']*10000,
                       ['Harry Jones', 'Mary Jones', 'Susan Jones']*10000])

%timeit df.values[:] = np.core.defchararray.replace(df.values.astype(str), ' ', '_')
10 loops, best of 3: 181 ms per loop

%timeit df.replace(' ', '_', regex=True)
1 loop, best of 3: 4.14 s per loop

%timeit df.stack().str.replace(' ', '_').unstack()
10 loops, best of 3: 99.2 ms per loop

Use replace method of dataframe:

df.replace('\s+', '_',regex=True,inplace=True)

pandas

stack / unstack with str.replace

df.stack().str.replace(' ', '_').unstack()

      Person_1    Person_2     Person_3
0   John_Smith  Jane_Smith   Mark_Smith
1  Harry_Jones  Mary_Jones  Susan_Jones

numpy

pd.DataFrame(
    np.core.defchararray.replace(df.values.astype(str), ' ', '_'),
    df.index, df.columns)

      Person_1    Person_2     Person_3
0   John_Smith  Jane_Smith   Mark_Smith
1  Harry_Jones  Mary_Jones  Susan_Jones

time testing