在数据库表中,我们找到具有文件路径的记录,处理文件的员工以及处理文件的时间戳。 表“日志”每位员工仅包含几千条记录。每位员工每天最多都有几条带有时间戳的记录(尽管不是唯一的时间戳)。 我想提取一个元组列表,其中包含一个日期和在该日期创建的表条目的数量。
我提供的代码可以工作,但是速度很慢。 2300条记录的22秒计算时间是荒谬的。 我已经将问题缩小到for循环中的“ logs_per_day =(query.select()。where(fn.date(cls.datetime)== checkday).count())”行中。 我了解在一个循环中执行许多查询可能并不好。此外,将datetime对象转换为日期可能也无济于事。 可以给我指出一种更好的方法吗?
import datetime
import os
from peewee import *
db = SqliteDatabase('logs.db')
# db = SqliteDatabase(':memory:')
now = datetime.datetime.now()
class BaseModel(Model):
class Meta:
database = db
class Log(BaseModel):
log_ID = AutoField()
datetime = DateTimeField()
letter = CharField()
disk_path = CharField()
ftp_path = CharField()
out = BooleanField()
employee = CharField(null=True)
class Meta:
table_name = 'log'
@classmethod
def get_histo_data(cls, employee="Some Dude", year=None):
"""returns a list with sublists (datetime object, integer)"""
if not year: # if no year was provided the query return all entries from the employee
query = cls.select().where(cls.employee == employee).order_by(cls.datetime)
print(employee, len(query), " entries")
firstday = query.order_by(cls.datetime).get().datetime.date()
lastday = query.order_by(cls.datetime.desc()).get().datetime.date()
else: # returns all entries in the given year
query = (cls
.select()
.where(cls.employee == employee, cls.datetime.year == year)
.order_by(cls.datetime)
)
print("{} has {} entries in the year {}".format(employee, len(query), year))
firstday = datetime.date(year, 1, 1)
lastday = datetime.date(year, 12, 31)
print("first day sent: ", firstday)
print("last day sent: ", lastday)
daydelta = (lastday-firstday).days
sendList = []
for i in range(daydelta+1): ### FIXME: This is extremely slow!!!
checkday = firstday + datetime.timedelta(days=i)
logs_per_day = (query
.select()
.where(fn.date(cls.datetime) == checkday)
.count()
)
# print(checkday, "*** logs that day: ", logs_per_day)
sendList.append([checkday, logs_per_day])
return sendList
def initialize():
db.connection()
db.create_tables([Log], safe=True)
db.close()
if __name__ == '__main__':
initialize()
Log.get_histo_data(employee="Mr Someone", year=2018)
输出应类似于“ [(2018-11-12,157),(2018-11-13,12),(2018-11-14,0)...]
答案 0 :(得分:-1)
自己发现:
def get_histo_data(cls, employee="Some dude", year=None):
"""returns a list of tuples (datetime object, integer)"""
if not year: # if no year was provided the query return all entries from the employee
query = cls.select().where(cls.employee == employee).order_by(cls.datetime)
print(employee, len(query), " entries")
firstday = query.order_by(cls.datetime).get().datetime.date()
lastday = query.order_by(cls.datetime.desc()).get().datetime.date()
else: # returns all entries in the given year
query = cls.get_query_by_year(employee, year)
print("{} has {} entries in the year {}".format(employee, len(query), year))
firstday = datetime.date(year, 1, 1)
lastday = datetime.date(year, 12, 31)
### count the entries
logDict = dict()
for record in query:
date = record.datetime.date()
if date not in logDict:
logDict[date] = 1
else:
logDict[date] += 1
### fill the null days
daydelta = (lastday-firstday).days
for i in range(daydelta+1):
checkday = firstday + datetime.timedelta(days=i)
if checkday not in logDict:
logDict[checkday] = 0
else:
continue
return list(sorted(logDict.items()))