CODEWITHSUNDEEP
pythonnumpysympy
George
George

Reputation: 5691

call lambdify in a loop, avoid explicitly call

I have this code:

var = ['a','b','c']
arr = np.array([ [1,2,3,4,5,6,7,8,9],
        [0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]
      ])

y = np.hsplit(arr,len(var))

newdict = {}
for idx,el in enumerate(y):
    newdict[str(var[idx])] = el

print(newdict)

I am splitting the array in order to have 3 new arrays, one for each variable in the var list.

Then , I am creating a new dictionary in order to assign to every variable the corresponding array.So, my result now is:

{'a': array([[ 1. ,  2. ,  3. ],
       [ 0.1,  0.2,  0.3]]), 'b': array([[ 4. ,  5. ,  6. ],
       [ 0.4,  0.5,  0.6]]), 'c': array([[ 7. ,  8. ,  9. ],
       [ 0.7,  0.8,  0.9]])}

Now, I have an expression to evaluate:

expr = sympify('a + b +c')
f = lambdify(var, expr, 'numpy')

result = f(newdict['a'], newdict['b'], newdict['c'])

print(result)

So , I am using lambdify and I am receiving the correct result:

[[ 12.   15.   18. ]
 [  1.2   1.5   1.8]]

My question is how to avoid calling explicitly f(newdict['a'], newdict['b'], newdict['c'])?

How can I have lambdify in a loop?

Upvotes: 3

Views: 202

Answers (1)

Kirill Bulygin
Kirill Bulygin

Reputation: 3836

For your particular commutative function (a + b + c):

f(*newdict.values())

For non-commutative functions, it's required to specify key order (as keys are unordered in a dict), e.g.:

f(*[v for _, v in sorted(newdict.items())])

With explicit keys:

f(*[newdict[k] for k in 'abc'])

Using OrderedDict:

from collections import OrderedDict
newdict = OrderedDict()
for idx,el in enumerate(y):
    newdict[str(var[idx])] = el

f(*newdict.values())

(f(*[1, 2, 3]) is equivalent to f(1, 2, 3).)

Upvotes: 4

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