【发布时间】:2020-06-02 18:33:44
【问题描述】:
样本数据
{"transaction": {"merchant": "merchantA", "amount": 20, "time": "2019-02-13T10:00:00.000Z"}}
{"transaction": {"merchant": "merchantB", "amount": 90, "time": "2019-02-13T11:00:01.000Z"}}
{"transaction": {"merchant": "merchantC", "amount": 90, "time": "2019-02-13T11:00:10.000Z"}}
{"transaction": {"merchant": "merchantD", "amount": 90, "time": "2019-02-13T11:00:20.000Z"}}
{"transaction": {"merchant": "merchantE", "amount": 90, "time": "2019-02-13T11:01:30.000Z"}}
{"transaction": {"merchant": "merchantE", "amount": 90, "time": "2019-02-13T11:02:30.000Z"}}
.
.
我有一些这样的代码
df = pd.DataFrame()
for line in sys.stdin:
data = json.loads(line)
# df1 = pd.DataFrame(data["transaction"], index=[len(df.index)])
df1 = pd.DataFrame(data["transaction"], index=[data['transaction']['time']])
df1['time'] = pd.to_datetime(df1['time'])
df = df.append(df1)
# df['count'] = df.rolling('2min', on='time', min_periods=1)['amount'].count()
print(df)
print(len(df[df.merchant.eq(data['transaction']['merchant']) & df.amount.eq(data['transaction']['amount'])].index))
电流输出
2019-02-13T10:00:00.000Z merchantA 20 2019-02-13 10:00:00
2019-02-13T11:00:01.000Z merchantB 90 2019-02-13 11:00:01
2019-02-13T11:00:10.000Z merchantC 90 2019-02-13 11:00:10
2019-02-13T11:00:20.000Z merchantD 90 2019-02-13 11:00:20
2019-02-13T11:01:30.000Z merchantE 90 2019-02-13 11:01:30
2019-02-13T11:02:30.000Z merchantE 90 2019-02-13 11:02:30
2
预期输出
2019-02-13T10:00:00.000Z merchantA 20 2019-02-13 10:00:00
2019-02-13T11:00:01.000Z merchantB 90 2019-02-13 11:00:01
2019-02-13T11:00:10.000Z merchantC 90 2019-02-13 11:00:10
2019-02-13T11:00:20.000Z merchantD 90 2019-02-13 11:00:20
2019-02-13T11:01:30.000Z merchantE 90 2019-02-13 11:01:30
由于数据正在流式传输。我想检查是否有重复记录(其商家和金额值相同)在两分钟内到达,所以我将其丢弃并且不对其进行处理。将其打印为副本。
我必须对索引压缩或 groupby 做些什么吗?但是然后如何等同于多列。 或者两列上有一些滚动条件,但找不到任何方法。
我在这里错过了什么?
谢谢
编辑
#dup = df[df.duplicated(subset=['merchant', 'amount'], keep=False)]
res = df.loc[(df.merchant == data['transaction']['merchant']) & (df.amount == data['transaction']['amount'])]
# res['timediff'] = pd.to_timedelta((data['transaction']['time'] - res['time']), unit='T')
res['timediff'] = (data['transaction']['time'] - res['time'])
if len(res.index) >1:
print(res)
所以我尝试这样的事情,如果结果小于 120 秒,我可以处理它。 但生成的df目前以
的形式 merchant amount time concat timediff
2019-02-13 11:03:00 merchantF 10 2019-02-13 11:03:00 merchantF10 -1 days +23:59:20
2019-02-13 11:02:20 merchantF 10 2019-02-13 11:02:20 merchantF10 00:00:00
2019-02-13 11:01:30 merchantE 10 2019-02-13 11:01:30 merchantE10 00:01:00
2019-02-13 11:02:00 merchantE 10 2019-02-13 11:02:00 merchantE10 00:00:30
2019-02-13 11:02:30 merchantE 10 2019-02-13 11:02:30 merchantE10 00:00:00
-1 天 +23:59:20 这种格式我觉得可以用绝对值代替?
如何将时间转换为可以与 120 秒比较的格式? pd.to_deltatime() 对我不起作用,或者我使用错误。
【问题讨论】:
-
你能添加预期的输出吗? DataFrame 看起来如何?可能是 minimal, complete, and verifiable example 的必要更改数据
-
@jezrael 我用当前和预期的输出编辑和更新了问题
标签: python sql pandas duplicates rolling-computation