说我有两个数据集基础和付款.
基数是：
[ id,timestamp,value]
付款是：
[ payment_id,value,gateway ]
我想对付基础付款.理想的结果是：
[id,时间戳,值,payment_id,网关,概率]
基本上,它应该告诉我给定基本条目最可能的payment_id.匹配应同时考虑日期时间和值.如果只给出最高的概率,我会感到满意,但是也不会因为提出第二/第三建议而困扰我.
到目前为止,我已经阅读了一些有关模糊匹配和相似性学习,余弦相似性和内容的内容,但是似乎无法将这些内容应用于我的问题.
我想到了手动执行以下操作：
for each_entry in base:
value_difference = base['value'] - payment['value']
time_difference = base['timestamp'] - payment['timestamp']
if value_difference <= 0.1 and time_difference <= 0.1:
#if the difference is small,then tell me the payment_id.
事实是,这看起来像是一种真正的“转储”方法,如果有多个与条件匹配的payment_entry可能会发生冲突,并且我将不得不手动微调参数以获得良好的结果.
我希望找到一种更智能,更自动的方式来协调这两个数据集.
有人对如何解决问题有任何建议吗？
编辑：我当前的状态：
import pandas as pd
import time
from itertools import islice
from pandas import ExcelWriter
import datetime
from random import uniform
orders = pd.read_excel("Orders.xlsx")
pmts = pd.read_excel("Payments.xlsx")
pmts['date'] = pd.to_datetime(pmts.date)
orders['data'] = pd.to_datetime(orders.data)
payment_list = []
for index,row in pmts.iterrows():
new_entry = {}
ts = row['date']
new_entry['id'] = row['id']
new_entry['date'] = ts.to_pydatetime()
new_entry['value'] = row['value']
new_entry['types'] = row['pmt']
new_entry['results'] = []
payment_list.append(new_entry)
order_list = []
for index,row in orders.iterrows():
new_entry = {}
ts = row['data']
new_entry['id'] = row['Id1']
new_entry['date'] = ts.to_pydatetime()
new_entry['value'] = row['valor']
new_entry['types'] = row['nome']
new_entry['results'] = []
order_list.append(new_entry)
for each_entry in order_list:
for each_payment in payment_list:
delta_value = (each_entry['value'] - each_payment['value'])
try:
delta_time = abs(each_entry['date'] - each_payment['date'])
except:
TypeError
pass
results = []
delta_ref = datetime.timedelta(minutes=60)
if delta_value == 0 and delta_time < delta_ref:
result_type = each_payment['types']
result_id = each_payment['id']
results.append(result_type)
results.append(delta_time)
results.append(result_id)
each_entry['results'].append(results)
result_id = each_entry['id']
each_payment['results'].append(result_id)
orders2 = pd.DataFrame(order_list)
writer = ExcelWriter('OrdersList.xlsx')
orders2.to_excel(writer)
writer.save()
pmts2 = pd.DataFrame(payment_list)
writer = ExcelWriter('PaymentList.xlsx')
pmts2.to_excel(writer)
writer.save()
好,现在我得到了一些东西.它向我返回所有具有相同值和小于x的时间增量(在这种情况下为60分钟)的条目.最好只给我最有可能的结果,也不会给出匹配正确的可能性(相同的数量,较短的时间范围).将继续尝试.
最佳答案
最简单的方法可能是选择差异最小的基本/付款对.例如：
base_data = [...] # all base data
payment_data = [...] # all payment data
def prop_diff(a,b,props):
# this iterates through all specified properties and
# sums the result of the differences
return sum([a[prop]-b[prop] for prop in props])
def join_data(base,payment):
# you need to implement your merging strategy here
return joined_base_and_payment
results = [] # where we will store our merged results
working_payment = payment_data.copy()
for base in base_data:
# check the difference between the lists
diffs = []
for payment in working_payment:
diffs.append(prop_diff(base,payment,['value','timestamp']))
# find the index of the payment with the minimum difference
min_idx = 0
for i,d in enumerate(diffs):
if d < diffs[min_idx]:
min_idx = i
# append the result of the joined lists
results.append(join_data(base,working_payment[min_idx]))
del working_payment[min_idx] # remove the selected payment
print(results)
基本思想是找到列表之间的总差异,并选择差异最小的对.在这种情况下,我复制了payment_data,所以我们不会破坏它,并且一旦我们将其与基本匹配并附加了结果,我们实际上就删除了该付款条目.