map in python2 vs python3

Question

I'm a beginner python user and I've ran the following code on both python2.7 and python3.4.3

import matplotlib.pyplot as plt
import numpy as np
import scipy.stats as stats

alpha = 1
n = 100

u = stats.uniform(0,1)
F_inverse = lambda u: 1/alpha*np.log(1/(1-u))
v = np.array(map(F_inverse, u.rvs(n)))
print(v)

fig, ax = plt.subplots(1,1)
stats.probplot(v, (1,), dist='expon', plot=ax)
plt.show()

On python2 i get a nice array like this:

array([  2.29133808e+00,   1.63236151e+00,   6.77776227e-01,
         3.33668250e-01,   1.77830890e+00,   3.06193068e-01,
         2.10677775e+00,   1.30525788e-01,   2.97056775e-01,
                           ...
         1.31463775e+00,   1.41840428e-03,   8.60594737e-01,
         1.80644880e-01])

On python3 i get this:

array(<map object at 0x7f8aab6f3ef0>, dtype=object)

If I change this:

v = np.array(map(F_inverse, u.rvs(n)))

to

v = list(map(F_inverse, u.rvs(n)))

it works fine on both but I would want to use an array instead. Is there a way to get this to work with np.array?

For the record, when only one iterator is involved, `list(map(foo, bar))` on Py3 is exactly equivalent to `map(foo, bar)` on Py2. So if the latter works somewhere on Py2, and it must work that way on Py3, substitute in the former, and it will be fine. — ShadowRanger, Feb 18 '17 at 02:21

falsetru · Accepted Answer · 2017-02-18T02:24:08.503

Convert the map object to list, then pass it to numpy.array.

v = np.array(list(map(F_inverse, u.rvs(n))))

Or use list comprehension instead of map to make a list instead of map object:

v = np.array([F_inverse(x) for x in u.rvs(n)])

But, you don't need to use map or list comprehension is not requierd; just calling F_inverse directly is enough because the F_inverse uses vectorized operations:

v = F_inverse(u.rvs(n))

map in python2 vs python3

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