進入一個SQLite數據庫

Question

我用下面的python3代碼在一個sqlite3的數據庫中添加和更新條目大量條目：

def increment_person_counts(count_per_person): 
    with sqlite3.connect(r'./people_database') as connection: 
     cursor = connection.cursor() 
     for person in count_per_person: 
     if cursor.execute('select * from personCounts where person = ?', [person]).fetchone()==None: 
      cursor.execute('insert into personCounts(person, count) values (?, ?)', [person, count_per_person[person]]) 
     else: 
      cursor.execute('update personCounts SET count=count + ? WHERE person=?', [count_per_person[person], person]) 
     connection.commit() 

count_per_person中包含4萬個條目，我似乎能夠添加/每秒更新大約100個條目，這意味着需要半天才能添加這些值。有沒有更好/更快的方法來做到這一點，我應該考慮？

感謝您的幫助，

巴里

Answer 1

你可以閱讀你的整個'select * from personCounts'成在開始一個python set()，然後檢查只是針對這一套。

def increment_person_counts(count_per_person): 
    with sqlite3.connect(r'./people_database') as connection: 
     cursor = connection.cursor() 
     cursor.execute('select person from personCounts') 
     known_persons = set(row[0] for row in cursor.fetchall()) 
     for person, count in count_per_person.iteritems(): 
     if person in known_persons: 
      cursor.execute('insert into personCounts(person, count) values (?, ?)', [person, count]) 
     else: 
      cursor.execute('update personCounts SET count=count + ? WHERE person=?', [count, person]) 
     connection.commit() 

UPDATE：我的意見後，這裏是executemany更新：

def increment_person_counts(count_per_person): 
    with sqlite3.connect(r'./people_database') as connection: 
     cursor = connection.cursor() 
     cursor.execute('select person from personCounts') 
     known_persons = set(row[0] for row in cursor.fetchall()) 
     cursor.executemany('insert into personCounts(person, count) values (?, ?)', ((person, count) for count_per_person.iteritems() if person in known_persons)) 
     for person, count in count_per_person.iteritems(): 
      if person not in known_persons: 
       cursor.execute('update personCounts SET count=count + ? WHERE person=?', [count, person]) 
     connection.commit()