熊猫时间序列事件之间的时间

Question

如何计算熊猫时间序列中"事件"之间的时间(天数)？例如,如果我有以下时间序列,我想在系列中的每一天知道自上次以来已经过了多少天TRUE

            event
2010-01-01  False
2010-01-02   True
2010-01-03  False
2010-01-04  False
2010-01-05   True
2010-01-06  False

我做的方式似乎过于复杂,所以我希望有更优雅的东西.显然,迭代遍历行的for循环可行,但我正在寻找理想的矢量化(可扩展)解决方案.我目前的尝试如下:

date_range = pd.date_range('2010-01-01', '2010-01-06')
df = pd.DataFrame([False, True, False, False, True, False], index=date_range, columns=['event'])
event_dates = df.index[df['event']]
df2 = pd.DataFrame(event_dates, index=event_dates, columns=['max_event_date'])
df = df.join(df2)
df['max_event_date'] = df['max_event_date'].cummax(axis=0, skipna=False)
df['days_since_event'] = df.index - df['max_event_date']

            event max_event_date  days_since_event
2010-01-01  False            NaT               NaT
2010-01-02   True     2010-01-02            0 days
2010-01-03  False     2010-01-02            1 days
2010-01-04  False     2010-01-02            2 days
2010-01-05   True     2010-01-05            0 days
2010-01-06  False     2010-01-05            1 days

Answer 1

继续改进这个答案,并希望有人采用'the'pythonic方式.在那之前,我认为最终的更新效果最好.

last = pd.to_datetime(np.nan)
def elapsed(row):
    if not row.event:
        return row.name - last
    else:
        global last
        last = row.name
        return row.name-last

df['elapsed'] = df.apply(elapsed,axis=1)

df
            event  elapsed
2010-01-01  False      NaT
2010-01-02   True   0 days
2010-01-03  False   1 days
2010-01-04  False   2 days
2010-01-05   True   0 days
2010-01-06  False   1 days

:::::::::::::

留下以前的答案,尽管它们不是最佳的

:::::::::

而不是进行多次传递,似乎更容易循环索引

df['elapsed'] = 0
for i in df.index[1:]:
    if not df['event'][i]:
        df['elapsed'][i] = df['elapsed'][i-1] + 1

::::::::::::

让我们说'特鲁斯'是你感兴趣的事件.

trues = df[df.event==True]
trues.Dates = trues.index #need this because .diff() doesn't work on the index
trues.Elapsed = trues.Dates.diff()