Move non-empty cells to the left in pandas DataFrame

Here's what I did:

I unstacked your dataframe into a longer format, then grouped by the name column. Within each group, I drop the NaNs, but then reindex to the full h1 thought h4 set, thus re-creating your NaNs to the right.

from io import StringIO
import pandas

def defragment(x):
    values = x.dropna().values
    return pandas.Series(values, index=df.columns[:len(values)])

datastring = StringIO("""\
Name    h1    h2    h3    h4
A       1     nan   2     3
B       nan   nan   1     3
C       1     3     2     nan""")

df = pandas.read_table(datastring, sep='\s+').set_index('Name')
long_index = pandas.MultiIndex.from_product([df.index, df.columns])

print(
    df.stack()
      .groupby(level='Name')
      .apply(defragment)
      .reindex(long_index)  
      .unstack()  
)

And so I get:

   h1  h2  h3  h4
A   1   2   3 NaN
B   1   3 NaN NaN
C   1   3   2 NaN

First, make function.

        def squeeze_nan(x):
            original_columns = x.index.tolist()

            squeezed = x.dropna()
            squeezed.index = [original_columns[n] for n in range(squeezed.count())]

            return squeezed.reindex(original_columns, fill_value=np.nan)

Second, apply the function.

df.apply(squeeze_nan, axis=1)

You can also try axis=0 and .[::-1] to squeeze nan to any direction.

[EDIT]

@Mxracer888 you want this?

def squeeze_nan(x, hold):
    if x.name not in hold:
        original_columns = x.index.tolist()

        squeezed = x.dropna()
        squeezed.index = [original_columns[n] for n in range(squeezed.count())]

        return squeezed.reindex(original_columns, fill_value=np.nan)
    else:
        return x

df.apply(lambda x: squeeze_nan(x, ['B']), axis=1)