Numpy.diff problem and numpy.cov problem in Python

Question

from matplotlib.finance import quotes_historical_yahoo_ochl as getData
import scipy as sp 
ticker='IBM' 
begdate=(2013,9,1) 
enddate=(2013,11,11) 
data= getData(ticker, begdate, enddate,asobject=True, adjusted=True) 
p=data.aclose 
d=sp.diff(p)
cov_=sp.cov(d[:-1],d[1:]) 
if cov_[0,1]<0: 
    print("Roll spread for ", ticker, 'is', round(2*sp.sqrt(-cov_[0,1]),3)) 
else: 
    print("Cov is positive for ",ticker, 'positive', round(cov_[0,1],3))

My effort is to calculate the above code in Python but since matplotlib.finance has been deprecated I tried another road using pandas_datareader.data.

Doing so I made this one :

import pandas_datareader.data as web
end = '2013-11-11'
start = '2013-09-01'
get_px = lambda x: web.DataReader(x, 'yahoo', start=start, end=end)['Adj Close']

symbols = ['IBM']
data = pd.DataFrame({sym:get_px(sym) for sym in symbols})
data = data.rename({'IBM': 'Adj Close'}, axis=1)
p1 = data.dropna()
p = np.asarray(p1);p
d = np.diff(p);d
cov_= np.cov(d[:-1],d[1:])
cov_

I receive an error

/opt/anaconda3/lib/python3.8/site-packages/numpy/lib/function_base.py:380: RuntimeWarning: Mean of empty slice.
  avg = a.mean(axis)
RuntimeWarning: Mean of empty slice.

if cov_[0,1]<0:
    print("Roll spread for ", symbols, 'is', round(2*sp.sqrt(-cov_[0,1]),3))
else:
    print("Cov is positive for ",symbols, 'positive', round(cov_[0,1],3)

What is my mistake ? I don’t know in the first code what type of object the data object is.And probably there is where the problem starts.

Updated

The line code

d = np.diff(p);d

is empty.The result is :

 array([], shape=(50, 0), dtype=float64)

The needed result is the 1.136

You are receiving a warning, not an error. From the output, we don't know where the warning is generated. Take a look here. https://stackoverflow.com/questions/5644836/in-python-how-does-one-catch-warnings-as-if-they-were-exceptions — sarema, Jan 18 '22 at 14:29
@sarema yes it is helpful your link in your comment but still does not produce the same results as that in the pic. — Homer Jay Simpson, Jan 18 '22 at 15:06
also the ```d = np.diff(p);d``` is empty.The result is : ```array([], shape=(50, 0), dtype=float64)``` — Homer Jay Simpson, Jan 18 '22 at 15:10
Please update your question with the actual output that you get, and the output you expect. — sarema, Jan 18 '22 at 15:16

StupidWolf · Accepted Answer · 2022-01-18T18:00:48.623

1

Using your code, the shape of p is wrong:

p.shape
Out[8]: (50, 1)

When you apply np.diff() onto this, it applies the difference on every entry, instead of across the vector as you intended, hence the error.

You can do something like this:

p = p1['Adj Close'].ravel()
d = np.diff(p)
cov_= np.cov(d[:-1],d[1:])
cov_

array([[ 3.12100146, -0.15854027],
       [-0.15854027,  3.20098506]])

Not very sure how you define the roll, I think:

np.sqrt(abs(cov_))
 
array([[1.76663563, 0.39817116],
       [0.39817116, 1.78912969]])

What you need to get 1.766 is:

np.sqrt(abs(cov_))[0,0]

edited Jan 18 '22 at 18:00

answered Jan 18 '22 at 17:46

StupidWolf

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it gives me "Roll spread for ['IBM'] is 0.796" and should be 1.136. – Homer Jay Simpson Jan 18 '22 at 17:59
see update, not sure how you define roll – StupidWolf Jan 18 '22 at 18:03
see the picture that I have added in the original post.I have removed it because they told that I must not upload photos of code – Homer Jay Simpson Jan 18 '22 at 18:10
I see, I am using `np.cov()` so the final code is correct – StupidWolf Jan 18 '22 at 18:12
Yes but the result is wrong – Homer Jay Simpson Jan 18 '22 at 18:13

Numpy.diff problem and numpy.cov problem in Python

1 Answers1