Python: Accessing The Proper Values In An Array

I want to calculate the following equation: calc = value_a(2D) - (value_b(0D) + value_b(1D))/10000 value_a(2D) corresponds to type **a**, year **2D** and value **1.1275** value_b(

Solution 1:

It looks like you really could use a multi-index here:

In [4]:df.reset_index(inplace=True)In [5]:dfOut[5]:typeyeardatevalue0a2D2015-02-09   1.12751b10M2015-02-09  58.12502b11M2015-02-09  68.37503b1M2015-02-09   3.34504b1W2015-02-09   0.89005b1Y2015-02-09  79.37506b2M2015-02-09   7.53507b2W2015-02-09   1.80008b3M2015-02-09  11.61009b3W2015-02-09   2.480010b4M2015-02-09  16.200011b5M2015-02-09  21.650012b6M2015-02-09  27.100013b7M2015-02-09  33.625014b8M2015-02-09  41.375015b9M2015-02-09  49.500016b0D2015-02-09   0.000017b1D2015-02-09   0.1250In [6]:df.set_index(['type','year'],inplace=True)In [7]:dfOut[7]:datevaluetypeyeara2D2015-02-09   1.1275b10M2015-02-09  58.125011M2015-02-09  68.37501M2015-02-09   3.34501W2015-02-09   0.89001Y2015-02-09  79.37502M2015-02-09   7.53502W2015-02-09   1.80003M2015-02-09  11.61003W2015-02-09   2.48004M2015-02-09  16.20005M2015-02-09  21.65006M2015-02-09  27.10007M2015-02-09  33.62508M2015-02-09  41.37509M2015-02-09  49.50000D2015-02-09   0.00001D2015-02-09   0.1250

Then simply:

In[8]: df.loc['a','2D'].value- (df.loc['b', '0D'].value + df.loc['b','1D'].value)/10000Out[8]: 1.1274875

Note, suppose I have multiple years (this I made by simply concatenating the df to itself):

In [24]:df2Out[24]:typeyeardatevalue0a2D2015-02-09   1.12751b10M2015-02-09  58.12502b11M2015-02-09  68.37503b1M2015-02-09   3.34504b1W2015-02-09   0.89005b1Y2015-02-09  79.37506b2M2015-02-09   7.53507b2W2015-02-09   1.80008b3M2015-02-09  11.61009b3W2015-02-09   2.480010b4M2015-02-09  16.200011b5M2015-02-09  21.650012b6M2015-02-09  27.100013b7M2015-02-09  33.625014b8M2015-02-09  41.375015b9M2015-02-09  49.500016b0D2015-02-09   0.000017b1D2015-02-09   0.125018a2D2015-02-10   1.127519b10M2015-02-10  58.125020b11M2015-02-10  68.375021b1M2015-02-10   3.345022b1W2015-02-10   0.890023b1Y2015-02-10  79.375024b2M2015-02-10   7.535025b2W2015-02-10   1.800026b3M2015-02-10  11.610027b3W2015-02-10   2.480028b4M2015-02-10  16.200029b5M2015-02-10  21.650030b6M2015-02-10  27.100031b7M2015-02-10  33.625032b8M2015-02-10  41.375033b9M2015-02-10  49.500034b0D2015-02-10   0.000035b1D2015-02-10   0.1250In [25]:df.iloc[-2,-1]=100000# this corresponds to (b, 0D) and used to be 0

As @cᴏʟᴅsᴘᴇᴇᴅ noted, you can group by the 'date' column:

In [26]: df2.groupby('date').apply(
     ...:     lambda df:
     ...:         df.loc['a','2D'].value
     ...:         - (df.loc['b', '0D'].value + df.loc['b','1D'].value)
     ...:         / 10000
     ...: )
Out[27]:
date
2015-02-091.1274872015-02-10   -8.872513
dtype: float64