I believe you need fillna for Series, then get indices of sorted values by argsort and last select by iloc - output is sorted columns:
print (df)
experiment_a experiment_b
0 EXPT_2011_06 NaN
1 EXPT_2010_06 NaN
2 NaN EXPT_2011_07
df = df.iloc[df['experiment_a'].fillna(df['experiment_b']).argsort()]
print (df)
experiment_a experiment_b
1 EXPT_2010_06 NaN
0 EXPT_2011_06 NaN
2 NaN EXPT_2011_07
Detail:
print (df['experiment_a'].fillna(df['experiment_b']))
0 EXPT_2011_06
1 EXPT_2010_06
2 EXPT_2011_07
Name: experiment_a, dtype: object
print (df['experiment_a'].fillna(df['experiment_b']).argsort())
0 1
1 0
2 2
Name: experiment_a, dtype: int64
I test more solutions, with np.where is a bit better performace, but mainly it depends of data:
print (df)
experiment_a experiment_b
0 EXPT_2011_03 NaN
1 NaN EXPT_2009_08
2 NaN EXPT_2010_06
3 EXPT_2010_07 NaN
4 NaN EXPT_2011_07
#[500000 rows x 2 columns]
df = pd.concat([df] * 100000, ignore_index=True)
In [41]: %timeit (df.iloc[(np.where(df['experiment_a'].isnull(), df['experiment_b'], df['experiment_a'])).argsort()])
1 loop, best of 3: 318 ms per loop
In [42]: %timeit (df.iloc[df['experiment_a'].fillna(df['experiment_b']).argsort()])
1 loop, best of 3: 335 ms per loop
In [43]: %timeit (df.iloc[df['experiment_a'].combine_first(df['experiment_b']).argsort()])
1 loop, best of 3: 333 ms per loop
In [44]: %timeit (df.iloc[df.experiment_a.where(df.experiment_a.notnull(), df.experiment_b).argsort()])
1 loop, best of 3: 342 ms per loop