CODEWITHSUNDEEP
pythonpandas
Chris
Chris

Reputation: 31206

pandas dataframe by index and integer

So I have a pandas dataframe indexed by date.

I need to grab a value from the dataframe by date...and then grab the value from the dataframe that was the day before...except I can't just subtract a day, since weekends and holidays are missing from the data.

It would be great if I could write:

x = dataframe.ix[date]

and 

i = dataframe.ix[date].index
date2 = dataframe[i-1]

I'm not married to this solution. If there is a way to get the date or index number exactly one prior to the date I know, I would be happy...(short of looping through the whole dataframe and testing to see if I have a match, and saving the count...)

Upvotes: 2

Views: 806

Answers (1)

EdChum
EdChum

Reputation: 394041

Use .get_loc to get the integer position of a label value in the index:

In [51]:
df = pd.DataFrame(index=pd.date_range(start=dt.datetime(2015,1,1), end=dt.datetime(2015,2,1)), data={'a':np.arange(32)})
df

Out[51]:
             a
2015-01-01   0
2015-01-02   1
2015-01-03   2
2015-01-04   3
2015-01-05   4
2015-01-06   5
2015-01-07   6
2015-01-08   7
2015-01-09   8
2015-01-10   9
2015-01-11  10
2015-01-12  11
2015-01-13  12
2015-01-14  13
2015-01-15  14
2015-01-16  15
2015-01-17  16
2015-01-18  17
2015-01-19  18
2015-01-20  19
2015-01-21  20
2015-01-22  21
2015-01-23  22
2015-01-24  23
2015-01-25  24
2015-01-26  25
2015-01-27  26
2015-01-28  27
2015-01-29  28
2015-01-30  29
2015-01-31  30
2015-02-01  31

Here using .get_loc on the index will return the ordinal position:

In [52]:    
df.index.get_loc('2015-01-10')

Out[52]:
9

pass this value using .iloc to get a row value by ordinal position:

In [53]:    
df.iloc[df.index.get_loc('2015-01-10')]

Out[53]:
a    9
Name: 2015-01-10 00:00:00, dtype: int32

You can then subtract 1 from this to get the previous row:

In [54]:     
df.iloc[df.index.get_loc('2015-01-10') - 1]

Out[54]:
a    8
Name: 2015-01-09 00:00:00, dtype: int32

Upvotes: 2

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