[英]Python pandas shift dataframe with time index value
I am quite new with python and am struggling with the shift
in pandas. 我对python很陌生,并且正在努力应对熊猫的
shift
。
I am comparing data, but it needs to be aligned to compare it. 我正在比较数据,但需要对其进行调整以进行比较。 To align the data, I only need to shift one of the data's index values.
为了对齐数据,我只需要移动数据的索引值之一。
Reference data: Data to be shifted:
acc acc
index index
1480681219**96**0000000 1 1480681220**04**0000000 8
1480681220**00**0000000 2 1480681220**08**0000000 9
1480681220**04**0000000 3 1480681220**12**0000000 7
1480681220**08**0000000 4 1480681220**16**0000000 10
1480681220**12**0000000 5 1480681220**20**0000000 6
(The bold editing option did not seem to work, but I wanted to highlight those parts of the indexes) (粗体编辑选项似乎无效,但是我想突出显示索引的那些部分)
I would like to shift my data frame with amount of extra time given. 我想在给定额外时间的情况下转移数据框。 Please note, the time is in nanoseconds.
请注意,时间以纳秒为单位。 I realized that something like
df.shift(2)
shifts my data 2 places, but I would like to shift my data with -80000000 nanoseconds which in this case is 2 places: 我意识到类似
df.shift(2)
数据会将我的数据移动2个位置,但是我想将数据移动-80000000纳秒,在这种情况下为2个位置:
Input: 输入:
acc
index
1480681220040000000 8
1480681220080000000 9
1480681220120000000 7
1480681220160000000 10
1480681220200000000 6
Desired output: 所需的输出:
acc
index
1480681219960000000 8
1480681220000000000 9
1480681220040000000 7
1480681220080000000 10
1480681220120000000 6
1480681220160000000 NaN
1480681220200000000 NaN
This is a smaller scale of my code: 这是我的代码的较小比例:
class device_data(object):
def __init__(self):
_index = [1480681220040000000,
1480681220080000000,
1480681220120000000,
1480681220160000000,
1480681220200000000]
self.df = pd.DataFrame({'acc': [8, 9, 7, 10, 6], 'index': _index})
self.df = self.df.set_index('index')
if __name__ == '__main__':
extratime = np.int64(-40000000)
session = dict()
session[2] = {'testnumber': '401',
'devicename': 'peanut'}
session[2]['data_in_device_class'] = device_data()
print session[2]['data_in_device_class'].df
if hasattr(session[2]['data_in_device_class'], 'df'):
session[2]['data_in_device_class'].df = session[2]['data_in_device_class'].df.shift(int(round(extratime)))
else:
pass
print session[2]['data_in_device_class'].df
When I ran the original code, it gave me this error: OverflowError: Python int too large to convert to C long
当我运行原始代码时,它给了我这个错误:
OverflowError: Python int too large to convert to C long
I used extratime = np.int64(extratime)
to solve the problem. 我用
extratime = np.int64(extratime)
解决了这个问题。 I notice that with the scaled down version of my code, that it is not really needed. 我注意到,随着代码的缩减版本,它并不是真正需要的。
My question still stands as how I could use shift to move my index with a value amount and not with the amount of places it needs to move? 我的问题仍然是,如何使用shift来移动索引值而不是需要移动的位数?
Thank you 谢谢
IIUC: IIUC:
You can just reassign your index with itself added to extra time. 您可以重新分配索引,而索引本身会增加额外的时间。
Consider the dataframe df
as an example 以数据帧
df
为例
df = pd.DataFrame(np.arange(100).reshape(5, -1))
df
I can "shift" the entire dataframe down like this 我可以像这样将整个数据框“下移”
df.index = df.index + 5
df
Let me know if this is on the mark. 让我知道这是否可行。 Otherwise, I'll delete it.
否则,我将其删除。
First you want to shift your index by the desired amount, and then reindex
, to make things easier I take a copy
here, shift the index, and we reindex
on the union
of the shifted index and the original index to introduce NaN
rows: 首先要通过所需量的指数移动,然后
reindex
,使事情变得更容易我拿一个copy
在这里,却将指数,我们reindex
对union
移位索引和原始索引引进NaN
行:
In [232]:
df1 = df.copy()
df1.index -= 80000000
df1.reindex(df1.index.union(df.index))
Out[232]:
acc
index
1480681219960000000 8.0
1480681220000000000 9.0
1480681220040000000 7.0
1480681220080000000 10.0
1480681220120000000 6.0
1480681220160000000 NaN
1480681220200000000 NaN
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