First of all, 1.79769313486e+308 is not the same as +inf. The former is the largest number which can be expressed with a 64-bit float, the latter is a special float.

If you just have very large numbers in your array, then:

A[A > 1e308] = 0

is sufficient. Thet'll replace oll elements above 1e308 with 0.

It is also possible to operate with the inf's. For example:

>>> fmax = np.finfo(np.float64).max
>>> pinf = float('+inf')
>>> ninf = float('-inf')
>>> fnan = float('nan')
>>> print fmax, pinf, ninf, fnan
1.79769313486e+308 inf -inf nan

So, these are completely different things. You may compare some of them:

>>> pinf > fmax
True
>>> ninf < 0.0
True
>>> pinf == pinf
True
>>> pinf == ninf
False

This looks good! However, nan acts differently:

>>> fnan > 0
False
>>> fnan < 0
False
>>> fnan == 0
False
>>> fnan < pinf
False
>>> fnan == fnan
False

You may use positive and negativi infinities with Numpy ndarraywithout any problems. This will work:

A[A == pinf] = 0.0

But if you have nans in the array, you'll get some complaints:

>>> np.array([fnan, pinf, ninf]) < 0
RuntimeWarning: invalid value encountered in less
[False, False, True]

So, it works but complains => do not use. The same without the nan:

>>> np.array([0.0, pinf, ninf]) < 0
[False, False, True]

If you want to do something with the nans (should you have them), use numpy.isnan:

A[np.isnan(A)] = 0.0

will change all nans into zeros.

And -- this you did not ask -- here is one to surprise your friends (*):

>>> [float('-0.0'), 0.0] * 3
[-0.0, 0.0, -0.0, 0.0, -0.0, 0.0]

Yep, float64 (and float32) have even a separate -0.0. In calculations it acts as an ordinary zero, though:

>>> float('-0.0') == 0.0
True

(*) Depending on the kind of people you call friends.

You can just replace NaN and infinite values with the following mask:

output[~np.isfinite(output)] = 0

>>> output
array([[1.        , 0.5       , 1.        , 1.        , 0.        ,
        1.        ],
       [1.        , 1.        , 0.5       , 1.        , 0.46524064,
        1.        ],
       [1.        , 1.        , 1.        , 0.        , 1.        ,
        1.        ]])

The easiest way to do this is to use numpy.ma.masked_invalid():

a = numpy.log(numpy.arange(15))
a.sum()
# -inf
numpy.ma.masked_invalid(a).sum()
# 25.19122118273868

You have to save the operation in your dataframe. One way is to use the parameter inplace=True:

df_fund['dly_retn'].replace(np.inf, 0, inplace=True)
na_fund['dly_retn'].replace(np.inf, 0, inplace=True)

Use numpy.delete(), which returns a new array with sub-arrays along an axis deleted.

numpy.delete(a, index)

For your specific question:

import numpy as np

a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
index = [2, 3, 6]

new_a = np.delete(a, index)

print(new_a)
# Output: [1, 2, 5, 6, 8, 9]

Note that numpy.delete() returns a new array since array scalars are immutable, similar to strings in Python, so each time a change is made to it, a new object is created. I.e., to quote the delete() docs:

"A copy of arr with the elements specified by obj removed. Note that delete does not occur in-place..."

If the code I post has output, it is the result of running the code.

To remove NaN values from a NumPy array x:

x = x[~numpy.isnan(x)]

Explanation

The inner function numpy.isnan returns a boolean/logical array which has the value True everywhere that x is not-a-number. Since we want the opposite, we use the logical-not operator ~ to get an array with Trues everywhere that x is a valid number.

Lastly, we use this logical array to index into the original array x, in order to retrieve just the non-NaN values.