1

我的数据看起来像：

[[datetime1, label1],
 [datetime2, label2],
 [datetime3, label3]]

标签是字符串。我有一个分箱参数 (delta)，它是 datetime.timedelta。

我正在尝试做的事情：

  1. 想出一组由 delta 等间隔的日期时间箱。换句话说，在下面， datetimebin2 - datetimebin1 = datetimebin3 - datetimebin2 = delta 。
  2. 将标签分箱到这些箱中。

所以我最终会得到类似的东西：

[[datetimebin1, [label1, label2],
 [datetimebin2, []],
 [datetimebin3, []],
 [datetimebin4, [label3]]

我被指向熊猫，但还没有找到我要找的东西。任何帮助深表感谢！

4

2 回答 2

3

我认为@DrV's 是正确答案，但我准备了一个示例，试图展示如何使用 Pandas 实现类似的事情：

import numpy
import pandas
import datetime
import time

# Binning delta

delta = datetime.timedelta(hours=1)

# Sample data

sample = [
    ['2014-08-09 16:30:00', 'label1'],
    ['2014-08-09 15:30:00', 'label2'],
    ['2014-08-09 14:30:00', 'label3'],
    ['2014-08-09 14:00:00', 'label4']
]

# Create dataframe and append UNIX timestamp column

df = pandas.DataFrame(sample)
df.columns = ['Datetime', 'Label']
df['Datetime'] = pandas.to_datetime(df['Datetime'])
df['UnixStamp'] = df['Datetime'].apply(lambda d: time.mktime(d.timetuple()))
df = df.set_index('Datetime')

# Calculate bins

bins = numpy.arange(min(df['UnixStamp']), max(df['UnixStamp']) + delta.seconds, delta.seconds)

# Group columns by datetime bin

def bin_from_tstamp(tstamp):

    diffs = [abs(tstamp - bin) for bin in bins]
    return bins[diffs.index(min(diffs))]

grouped = df.groupby(df['UnixStamp'].map(
    lambda t: datetime.datetime.fromtimestamp(bin_from_tstamp(t))
))

此时grouped包含按日期时间箱分组的数据集。

以下是打印的结果grouped.groups（其中键是日期时间箱，值是分组的日期时间）：

{
    numpy.datetime64('2014-08-09T18:00:00.000000000+0200'): [
        Timestamp('2014-08-09 16:30:00')
    ], 
    numpy.datetime64('2014-08-09T17:00:00.000000000+0200'): [
        Timestamp('2014-08-09 15:30:00')
    ], 
    numpy.datetime64('2014-08-09T16:00:00.000000000+0200'): [
        Timestamp('2014-08-09 14:30:00'), 
        Timestamp('2014-08-09 14:00:00'
    ]
}
于 2014-08-09T16:33:02.567 回答
2

这些方面应该做的事情：

# data: a lists of lists (length 2) of measurements
# res: resulting list of lists
# delta: time delta

# output list (will be a list of lists, as in the question

res = []
# end of first bin:
binstart = data[0][0]
res.append([binstart, []])

# iterate through the data item
for d in data:
    # if the data item belongs to this bin, append it into the bin
    if d[0] < binstart + delta:
        res[-1][1].append(d[1])
        continue

    # otherwise, create new empty bins until this data fits into a bin
    binstart += delta
    while d[0] > binstart + delta:
        res.append([binstart, [])
        binstart += delta

    # create a bin with the data
    res.append([binstart, [d[1]]])
于 2014-08-09T15:47:08.637 回答