Pandas DataFrame 如何查询最近的日期时间索引？
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Pandas DataFrame How to query the closest datetime index?
提问by Bryan Fok
How do i query for the closest index from a Pandas DataFrame? The index is DatetimeIndex
如何从 Pandas DataFrame 查询最近的索引？索引是 DatetimeIndex
2016-11-13 20:00:10.617989120 7.0 132.0
2016-11-13 22:00:00.022737152 1.0 128.0
2016-11-13 22:00:28.417561344 1.0 132.0
I tried this:
我试过这个：
df.index.get_loc(df.index[0], method='nearest')
but it give me InvalidIndexError: Reindexing only valid with uniquely valued Index objects
但它给了我 InvalidIndexError: Reindexing only valid with uniquely valued Index objects
Same error if I tried this:
如果我试过这个，同样的错误：
dt =datetime.datetime.strptime("2016-11-13 22:01:25", "%Y-%m-%d %H:%M:%S")
df.index.get_loc(dt, method='nearest')
But if I remove method='nearest'it works, but that is not I want, I want to find the closest index from my query datetime
但是如果我删除method='nearest'它有效，但这不是我想要的，我想从我的查询日期时间中找到最接近的索引
回答by jezrael
It seems you need first get position by get_locand then select by []:
看来你需要先得到位置get_loc，然后选择[]：
dt = pd.to_datetime("2016-11-13 22:01:25.450")
print (dt)
2016-11-13 22:01:25.450000
print (df.index.get_loc(dt, method='nearest'))
2
idx = df.index[df.index.get_loc(dt, method='nearest')]
print (idx)
2016-11-13 22:00:28.417561344
#if need select row to Series use iloc
s = df.iloc[df.index.get_loc(dt, method='nearest')]
print (s)
b 1.0
c 132.0
Name: 2016-11-13 22:00:28.417561344, dtype: float64
回答by Bryan Fok
I believe jezrael solution works, but not on my dataframe (which i have no clue why). This is the solution that I came up with.
我相信 jezrael 解决方案有效，但不适用于我的数据框（我不知道为什么）。这是我想出的解决方案。
from bisect import bisect #operate as sorted container
timestamps = np.array(df.index)
upper_index = bisect(timestamps, np_dt64, hi=len(timestamps)-1) #find the upper index of the closest time stamp
df_index = df.index.get_loc(min(timestamps[upper_index], timestamps[upper_index-1],key=lambda x: abs(x - np_dt64))) #find the closest between upper and lower timestamp
