Use Updated Value In Array

Question

This is the code that I am working with:

import pandas as pd
z = pd.Series(data = [1,2,3,4,5,6,7], index = xrange(1,8))
array = []
for i in range(1,8):
    array.append(z[i]*2)
    print array

It does exactly what I tell it to because I can't figure out how to do a simple iteration. This is the printed output

[2]
[2, 4]
[2, 4, 6]
[2, 4, 6, 8]
[2, 4, 6, 8, 10]
[2, 4, 6, 8, 10, 12]
[2, 4, 6, 8, 10, 12, 14]

What I want is for python to use the updated value in array so the desired output would be:

[2]
[2, 4]
[2, 4, 8]
[2, 4, 8, 16]
[2, 4, 8, 16, 32]
[2, 4, 8, 16, 32, 64]
[2, 4, 8, 16, 32, 64, 128]

Thank you for your help. Edit The example I first used was too simple so please answer using the example code below

import pandas as pd
sample = pd.Series(data = [ -3.2 ,  30.66,   7.71,   9.87], index = range(0,4))
testarray = []
for i in range(0,4):
    testarray.append(100000*(1+sample.values[i]/100))
    print testarray

This produces

[96800.0, 130660.0, 107710.0, 109870.0]

When the desired numbers are: 96800 126478.88 136230.4016 149676.3423

So instead of it using 100000 I want it to use 96800 for the second iteration and so on. Thank you!

hasumedic · Accepted Answer · 2016-02-16 01:38:08Z

4

I think that what you're trying to do is to compute powers of two, instead of multiplications of two:

array.append(2**z[i])

answered Feb 16, 2016 at 1:38

hasumedic

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Comments

hpaulj · Accepted Answer · 2016-02-16 02:28:00Z

2

You want to use the last value of the accumulation list in the expression.

For example:

data = [ -3.2 ,  30.66,   7.71,   9.87]  # the input list for `sample`

In [686]: test=[100000]    
In [687]: for i in range(4):
    test.append(test[-1]*(1+data[i]/100))
In [688]: test
Out[688]: [100000, 96800.0, 126478.88, 136230.401648, 149676.3422906576]

I could start this with test=[], but then I'd have test whether the list was empty, and use 1000000 instead of test[-1]. So putting the 100000 in the list to start with is logically simpler.

Another option is to maintain a temporary variable that is updated each iteration:

In [689]: mult=100000
In [690]: test=[]
In [691]: for i in range(4):
    test.append(mult*(1+data[i]/100))
    mult=test[-1]

Or

mult *= (1+data[i]/100)
test.append(mult)

But since this is pandas I could also do the calculation with one vectorized call. The numpy array equivalent is:

In [697]: data_arr=np.array(data)
In [698]: np.cumprod(1+data_arr/100)
Out[698]: array([ 0.968     ,  1.2647888 ,  1.36230402,  1.49676342])

cumprod is the cumulative product (like the more common cumulative sum).

Your first example can be produced with:

In [709]: np.cumprod([2 for _ in range(7)])
Out[709]: array([  2,   4,   8,  16,  32,  64, 128])

In [710]: np.cumprod(np.ones(7,int)*2)
Out[710]: array([  2,   4,   8,  16,  32,  64, 128])

edited Feb 16, 2016 at 2:28

answered Feb 16, 2016 at 2:14

hpaulj

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1 Comment

Adam Warner Over a year ago

can you please include what you set data equal to in the first part of your anwer

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