как использовать питона список понимание/словарь для печати каждого столбца в качестве уникальной переменной

Question

Предположим, у нас есть файл CSV

PROPERTY_ID,CLIENT_ID,FROM_YEAR 
1,5,2015 
2,6,2015 
3,9,2015 
4,9,2015 

Я пытаюсь передать каждое уникальное сочетание CLIENT_ID, PROPERTY_ID, FROM_YEAR в словарь или список , так что я могу поместить каждую пару "PROPERTY_ID, CLIENT_ID, FROM_YEAR" в запросах MySQL:

SELECT * FROM client_5 WHERE PROPERTY_ID = 1 and FROM_YEAR = 2015; 

SELECT * FROM client_6 WHERE PROPERTY_ID = 2 and FROM_YEAR = 2015; 

SELECT * FROM client_9 WHERE PROPERTY_ID = 3 and FROM_YEAR = 2015; 

SELECT * FROM client_9 WHERE PROPERTY_ID = 4 and FROM_YEAR = 2015; 

с переменной точки зрения:

1st round: 
$CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=5,1,2015 

2nd round 
$CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=6,2,2015 

3rd round 
$CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=9,3,2015 

4th round 
$CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=9,4,2015 

Я пытался использовать список понимание:

df = pd.read_csv("test.csv") 

df2=df.apply(tuple, 1).unique().tolist() 

for CLIENT_ID in [x[0] for x in df2]: 

    CLIENT_ID=CLIENT_ID.astype('str') 

    print "SELECT * FROM client"+CLIENT_ID 

    for PROPERTY_CODE in [y[1] for y in df2]: 

     PROPERTY_CODE=PROPERTY_CODE.astype('str') 

     print "WHERE PROPERTY_ID = "+PROPERTY_CODE 

возвращает следующее, что не мы ищем:

SELECT * FROM client_5 
WHERE FK_PROPERTY_ID = 1 
WHERE FK_PROPERTY_ID = 2 
WHERE FK_PROPERTY_ID = 3 
WHERE FK_PROPERTY_ID = 4 

Может кто просветит? Благодарю.

Answer 1

Это будет работать для вас: -

import csv 

with open('fileName.csv') as f: 
    reader = csv.reader(f) 
    next(reader, None) 
    for row in reader: 

     #print row 
     print """SELECT * FROM client_%s WHERE PROPERTY_ID = %s and FROM_YEAR = %s;"""%(row[1],row[0],row[2]) 

Answer 2

Я хотел бы использовать format

fstr = '$CLIENT_ID,$PROPERTY_ID,$FROM_YEAR={CLIENT_ID},{PROPERTY_ID},{FROM_YEAR}' 
df.drop_duplicates().apply(lambda x: fstr.format(**x), 1) 

0 $CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=5,1,2015 
1 $CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=6,2,2015 
2 $CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=9,3,2015 
3 $CLIENT_ID,$PROPERTY_ID,$FROM_YEAR=9,4,2015 
dtype: object 

Answer 3

Я думаю, что вы можете использовать apply с set и list:

L = list(set(df.apply(lambda x: 'SELECT * FROM client_{} WHERE PROPERTY_ID = {} and FROM_YEAR = {};'.format(x['CLIENT_ID'], x['PROPERTY_ID'], x['FROM_YEAR']),1))) 

print (L) 
['SELECT * FROM client_5 WHERE PROPERTY_ID = 1 and FROM_YEAR = 2015;', 
'SELECT * FROM client_9 WHERE PROPERTY_ID = 3 and FROM_YEAR = 2015;', 
'SELECT * FROM client_9 WHERE PROPERTY_ID = 4 and FROM_YEAR = 2015;', 
'SELECT * FROM client_6 WHERE PROPERTY_ID = 2 and FROM_YEAR = 2015;'] 

Answer 4

Это легко реализовать с помощью метода .format:

import pandas as pd 

df = pd.read_csv('test.csv') 
rows = df.apply(tuple, 1).unique().tolist() 

for (prop_id, client_id, year) in rows: 
    print("SELECT * FROM client_{client_id} WHERE property_id = {prop_id} AND from_year = {year}".format(
     prop_id=prop_id, 
     client_id=client_id, 
     year=year 
    )) 

В Python 3.6 вы можете использовать интерполяцию строки:

for (prop_id, client_id, year) in rows: 
    print(f"SELECT * FROM client_{client_id} WHERE property_id = {prop_id} AND from_year = {year}")