进入一个SQLite数据库

Question

我用下面的python3代码在一个sqlite3的数据库中添加和更新条目大量条目：

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4万个条目，我似乎能够添加/每秒更新大约100个条目，这意味着需要半天才能添加这些值。有没有更好/更快的方法来做到这一点，我应该考虑？

感谢您的帮助，

巴里

Answer 1

你可以阅读你的整个'select * from personCounts'成在开始一个python set()，然后检查只是针对这一套。

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：我的意见后，这里是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()