Question

是否可以使用pandas.cut制作datetime邮票中的邮箱？

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)

引发以下错误：

---------------------------------------------------------------------------
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'

Answer 1

这是我的解决方法。您可能需要稍微更改代码以满足您的精度需求。我使用日期作为例子如下：

# 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转换为数字距离度量，然后剪切/ qcut它。

Answer 2

老问题，但是对于任何未来的访问者，我认为这是计算浮点时间值以使用剪切的更清晰的方法：

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点值，我们可以在

上使用pd.cut()

是否可以在日期时间集上使用剪切？

2 个答案: