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Python's 'set' operator doesn't work with numpy.nan


Python's 'set' operator doesn't work with numpy.nan

By : jpeterson
Date : November 23 2020, 03:01 PM
will be helpful for those in need NaNs in a float64 array don't point to the same space in memory as np.NaN, (they, like every other number in the array, 8 bytes in the array). We can see this when we take the id:
code :
In [11]: x_numeric
Out[11]:
0   NaN
1   NaN
Name: a, dtype: float64

In [12]: x_numeric.apply(id)
Out[12]:
0    4657312584
1    4657312536
Name: a, dtype: int64

In [13]: id(np.nan)
Out[13]: 4535176264

In [14]: id(np.nan)
Out[14]: 4535176264
In [21]: s = set([np.nan])

In [22]: np.nan in s
Out[22]: True

In [23]: x_numeric.apply(lambda x: x in s)
Out[23]:
0    False
1    False
Name: a, dtype: bool
In [24]: np.nan == np.nan
Out[24]: False


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Why doesnt Pythons += (plus equals) operator modify variables from inner functions?

Why doesnt Pythons += (plus equals) operator modify variables from inner functions?


By : user3332349
Date : March 29 2020, 07:55 AM
Hope this helps The problem is that when you assign to a variable name inside a function, Python assumes you're trying to create a new local variable that will mask a similarly-named variable in outer scope. Since += has to get the value of mylist before it can modify it, it complains, because the local version of mylist isn't yet defined. MRAB's answer gives a clear explanation of the semantics.
On the other hand, when you do mylist.__iadd__([1]), you aren't assigning a new variable name inside the function. You're just using a built-in method to modify an already-assigned variable name. As long as you don't try to assign a new value to mylist, you won't have a problem. For the same reason, the line mylist[0] = 5 would also work inside inner if the definition of mylist in outer were mylist = [1].
code :
>>> outer()
>>> def outer():
...     mylist = []
...     def inner():
...         mylist.__iadd__([1])
...         mylist = []
...     inner()
... 
>>> outer()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 6, in outer
  File "<stdin>", line 4, in inner
UnboundLocalError: local variable 'mylist' referenced before assignment
>>> mylist = []
>>> def inner():
...     global mylist
...     mylist += [1]
... 
>>> inner()
>>> mylist
[1]
def outer():
    mylist = []
    def inner():
        nonlocal mylist
        mylist += [1]
    inner()
    print(mylist)
outer()
C++ operator overload doesnt work

C++ operator overload doesnt work


By : Malang Nyassi
Date : March 29 2020, 07:55 AM
Does that help The problem here is you need to learn what references are and the difference between std::ostream and std::ostream& is.
std::ostream& operator<< (std::ostream& stream, const CStudent& student)
Linq Contains operator doesnt work

Linq Contains operator doesnt work


By : user2090299
Date : March 29 2020, 07:55 AM
should help you out The JoinAlias is a part of QueryOver syntax. For Contains() we should use this:
code :
return summary
    .JoinAlias(s => s.District, () => district
                  , NHibernate.SqlCommand.JoinType.LeftOuterJoin)
    .WhereRestrictionOn(() => district.Name) // here we say what to restrict
        .IsLike("ALEK", MatchMode.Anywhere)  // and we say to use LIKE
    ;                                        // matching as: %ALEK%
Numpy generate array from csv - doesnt work with only 1 value in csv

Numpy generate array from csv - doesnt work with only 1 value in csv


By : user3382372
Date : March 29 2020, 07:55 AM
Any of those help I have a problem with importing a csv using numpy, it works fine when there are more than 1 values in the csv, e.g. "one,two,three" but if I have just one value in the csv it won't work - "one" , It's not the nicest code but it should work
code :
terms = [x for x in open(file_path,'r').read().replace('\n',',').split(',') if len(x)>0]
Why does Pythons "any()" function work on numpy array?

Why does Pythons "any()" function work on numpy array?


By : adies
Date : March 29 2020, 07:55 AM
wish of those help For one-dimensional arrays it works because the built-in Python-any-function just requires an iterable with items that can be cast to bools (and a one-dimensional array satisfies these conditions) but for multidimensional arrays it won't work:
code :
>>> import numpy as np

>>> any(np.ones((10, 10)))
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

>>> np.any(np.ones((10, 10)))
True
In [0]: arr = np.zeros((1000))

In [1]: %timeit any(arr)     
Out[1]: 215 µs ± 4.29 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [2]: %timeit np.any(arr)  
Out[2]: 31.2 µs ± 1.41 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
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