CODEWITHSUNDEEP
pythonlistpython-3.xenumerate
VlS
VlS

Reputation: 586

Enumerate function

Can someone explain this to me.

So, I have a list s with numbers from -1 to 1 and I want to extract position of certain numbers in a list.

Example:

s= [-1, 0.5, 0.2, -0.9]

z = enumerate(s)
y1 = []
for i,j in z:
    if j<=-0.8 and j>=-1:
        k = i
        y1.append(k)

y2 = []
for i,j in z:
    if j<=0.8 and j>=-0.8:
        k = i
        y2.append(k)

I get y1 = [0, 3] and y2 = []

But if I define second enumerate:

z1 = enumerate(s)
y1 = []
for i,j in z1:
    if j<=-0.8 and j>=-1:
        k = i
        y1.append(k)

z2 = enumerate(s)
y2 = []
for i,j in z2:
    if j<=0.8 and j>=-0.8:
        k = i
        y2.append(k)

I get result y1 = [0, 3] and y2 = [1, 2]

Why do I need second enumerate?

Upvotes: 3

Views: 710

Answers (3)

Iron Fist
Iron Fist

Reputation: 10951

enumerate is a lazy generator, once called and traversed all the way it's done. So, I think it's better you include its call directly in the for loop:

s= [-1, 0.5, 0.2, -0.9]

y1 = []
for i,j in enumerate(s):
    if -1 <= j <= -0.8: 
        y1.append(i)

y2 = []
for i,j in enumerate(s):
    if -0.8 <= j <= 0.8:
        y2.append(i)

A side note:

if j<=-0.8 and j>=-1: can be replaced by if -1 <= j <= -0.8:

Upvotes: 2

YBathia
YBathia

Reputation: 902

From python docs, the implementation of enumerate is as follow: 

def enumerate(collection):
    'Generates an indexed series:  (0,coll[0]), (1,coll[1]) ...'     
    i = 0
    it = iter(collection)
    while 1:
        yield (i, it.next())
        i += 1

Since it uses 'yield', it is a generator which stores the state when you iterate through it once in line 

for i,j in z

To use only one enumerate, convert the values of the generator to a list and reuse the list :

s= [-1, 0.5, 0.2, -0.9]

z = list(enumerate(s))

y1 = []

for i,j in z:
    if j<=-0.8 and j>=-1:
        k = i
        y1.append(k)

y2 = []
for i,j in z:
    if j<=0.8 and j>=-0.8:
        k = i
        y2.append(k)

print(y1)
print(y2)

Output:

[0, 3]
[1, 2]

Upvotes: 2

awesoon
awesoon

Reputation: 33691

enumerate returns an iterator to the sequence. Once traversed, it cannot be used again:

In [1]: l = [1, 2, 3, 4, 5]

In [2]: e = enumerate(l)

In [3]: e
Out[3]: <enumerate at 0x7fad7aea25e8>

In [4]: list(e)
Out[4]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]

In [5]: list(e)
Out[5]: []

You could wrap it with a list:

In [6]: e = list(enumerate(l))

In [7]: e
Out[7]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]

In [8]: list(e)
Out[8]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]

In [9]: list(e)
Out[9]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]

You could also use list comprehensions in order to simplify your solution:

y1 = [i for i, x in enumerate(s) if -1 <= x <= -0.8]
y2 = [i for i, x in enumerate(s) if -0.8 <= x <= 0.8]

Upvotes: 3

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