для цикла, использующего iterrows в пандах

Вы можете использовать первое cross соединение с merge а затем фильтровать значения с помощью boolean indexing . Последний удалить ненужные столбцы:

 data1['tmp'] = 1 data2['tmp'] = 1 df = pd.merge(data1, data2, on='tmp', how='outer') df = df[(df.lowerbound_address <= df.address) & (df.upperbound_address >= df.address)] df = df.drop(['lowerbound_address','upperbound_address', 'tmp'], axis=1) print (df) id address place 1 1 11123451 Y 2 2 78947591 X 

Другое решение с itertuples , последнее создание DataFrame.from_records :

 places = [] for row1 in data1.itertuples(): for row2 in data2.itertuples(): #print (row1.address) if (row2.lowerbound_address <= row1.address <= row2.upperbound_address): places.append((row1.id, row1.address, row2.place)) print (places) [(1, 11123451, 'Y'), (2, 78947591, 'X')] df = pd.DataFrame.from_records(places) df.columns=['id','address','place'] print (df) id address place 0 1 11123451 Y 1 2 78947591 X 

Другое решение:

 def f(x): for row2 in data2.itertuples(): if (row2.lowerbound_address <= x <= row2.upperbound_address): return pd.Series([x, row2.place], index=['address','place']) df = data1.set_index('id')['address'].apply(f).reset_index() print (df) id address place 0 1 11123451 Y 1 2 78947591 X 

РЕДАКТИРОВАТЬ:

Сроки :

N = 1000 :

Если значения saome не находятся в зоне действия, в решениях b и c нет. Проверьте последнюю строку df1 .

 In [73]: %timeit (data1.set_index('id')['address'].apply(f).reset_index()) 1 loop, best of 3: 2.06 s per loop In [74]: %timeit (a(df1a, df2a)) 1 loop, best of 3: 82.2 ms per loop In [75]: %timeit (b(df1b, df2b)) 1 loop, best of 3: 3.17 s per loop In [76]: %timeit (c(df1c, df2c)) 100 loops, best of 3: 2.71 ms per loop 

Код для таймингов :

 np.random.seed(123) N = 1000 data1 = pd.DataFrame({'id':np.arange(1,N+1), 'address': np.random.randint(N*10, size=N)}, columns=['id','address']) #add last row with value out of range data1.loc[data1.index[-1]+1, ['id','address']] = [data1.index[-1]+1, -1] data1 = data1.astype(int) print (data1.tail()) data2 = pd.DataFrame({'lowerbound_address':np.arange(1, N*10,10), 'upperbound_address':np.arange(10,N*10+10, 10), 'place': np.random.randint(40, size=N)}) print (data2.tail()) df1a, df1b, df1c = data1.copy(),data1.copy(),data1.copy() df2a, df2b ,df2c = data2.copy(),data2.copy(),data2.copy() 

 def a(data1, data2): data1['tmp'] = 1 data2['tmp'] = 1 df = pd.merge(data1, data2, on='tmp', how='outer') df = df[(df.lowerbound_address <= df.address) & (df.upperbound_address >= df.address)] df = df.drop(['lowerbound_address','upperbound_address', 'tmp'], axis=1) return (df) 

 def b(data1, data2): places = [] for row1 in data1.itertuples(): for row2 in data2.itertuples(): #print (row1.address) if (row2.lowerbound_address <= row1.address <= row2.upperbound_address): places.append((row1.id, row1.address, row2.place)) df = pd.DataFrame.from_records(places) df.columns=['id','address','place'] return (df) 

 def f(x): #use for ... else for add NaN to values out of range #http://stackoverflow.com/q/9979970/2901002 for row2 in data2.itertuples(): if (row2.lowerbound_address <= x <= row2.upperbound_address): return pd.Series([x, row2.place], index=['address','place']) else: return pd.Series([x, np.nan], index=['address','place']) 

 def c(data1,data2): data1 = data1.sort_values('address') data2 = data2.sort_values('lowerbound_address') df = pd.merge_asof(data1, data2, left_on='address', right_on='lowerbound_address') df = df.drop(['lowerbound_address','upperbound_address'], axis=1) return df.sort_values('id') print (data1.set_index('id')['address'].apply(f).reset_index()) print (a(df1a, df2a)) print (b(df1b, df2b)) print (c(df1c, df2c)) 

Только решение c с merge_asof работает с большим DataFrame :

N=1M :

 In [84]: %timeit (c(df1c, df2c)) 1 loop, best of 3: 525 ms per loop 

Подробнее о слиянии asof в документах .