上使用
是否可以使用pandas.cut以datetime邮票制作垃圾箱?
以下代码:
import pandas as pd
import StringIO
contenttext = """Time,Bid
2014-03-05 21:56:05:924300,1.37275
2014-03-05 21:56:05:924351,1.37272
2014-03-05 21:56:06:421906,1.37275
2014-03-05 21:56:06:421950,1.37272
2014-03-05 21:56:06:920539,1.37275
2014-03-05 21:56:06:920580,1.37272
2014-03-05 21:56:09:071981,1.37275
2014-03-05 21:56:09:072019,1.37272"""
content = StringIO.StringIO(contenttext)
df = pd.read_csv(content, header=0)
df['Time'] = pd.to_datetime(df['Time'], format='%Y-%m-%d %H:%M:%S:%f')
pd.cut(df['Time'], 5)
抛出以下错误:
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-3-f5387a84c335> in <module>()
16 df['Time'] = pd.to_datetime(df['Time'], format='%Y-%m-%d %H:%M:%S:%f')
17
---> 18 pd.cut(df['Time'], 5)
/home/???????/sites/varsite/venv/local/lib/python2.7/site-packages/pandas/tools/tile.pyc in cut(x, bins, right, labels, retbins, precision, include_lowest)
80 else:
81 rng = (nanops.nanmin(x), nanops.nanmax(x))
---> 82 mn, mx = [mi + 0.0 for mi in rng]
83
84 if mn == mx: # adjust end points before binning
TypeError: unsupported operand type(s) for +: 'Timestamp' and 'float'
老问题，但对于任何未来的访问者，我认为这是一个更清晰的方法来计算float timedelta使用cut:
import pandas as pd
import datetime as dt
# Get Days Since Date
today = dt.date.today()
df['days ago'] = (today - df['time']).dt.days
# Get Seconds Since Datetime
now = dt.datetime.now()
df['seconds ago'] = (now - df['time']).dt.seconds
# Minutes Since Datetime
# (no dt.minutes attribute, so we use seconds/60)
now = dt.datetime.now()
df['minutes ago'] = (now - df['times']).dt.seconds/60
所有这些列现在都是浮动值我们可以在
pd.cut()这是我的解决方案。您可能需要稍微更改代码以满足您的精度需求。我用date作为下面的例子:
# map dates to timedelta
today=dt.date.today()
# x below is a timedelta,
# use x.value below if you need more precision
df['days']=map(lambda x : x.days, df.Time - today)
pd.cut(df.days, bins=5)
有效地将datetime或date转换为数值距离度量，然后cut/qcut它