The problem with the python generator, boss, help.

g=(i for i in range(4))
for i in [1,10]:
    g=(i+j for j in g)
print(list(g))

may I ask why the result is: [20,21,22,23]
instead of: [11mem12121314]?

Python
Apr.09,2021
The characteristic of the

generator is that it does not execute immediately, but remembers the 'mode of production' until it is called.
in your example:
g = (i for i in range (4))-sharp at this time, if called by list, g will be called, but it will not be called, but in the generator
the following loop:
for i in [1jue 10]: 

g=(i+j for j in g)

does not mean that when the first cycle I is 1, g should be [1 br 2]. In fact, g is not called and all is not executed, just remember that the value of the generator is
the second cycle, I is 10, which is also not executed, but just remember the value of the variable I in the loop when it is called and executed by the print (list (g)) command. It's already 10.
so each value in the final g is executed twice in a row, that is, iExamj
, so the output of print (list (g)) is [20,21,22,23]

.

what's even more amazing is your code:
g = (i for i in range (4))
for i in [1Jing 10]: 

g=(i+j for j in g)

for i in g: 

print(i)-sharp20,41,84,171

isn't it weirder that the output is different from list (g)?
but if you change the name of the loop variable:
for k in g: 

print(k)-sharp20, 21, 22, 23

the output is consistent with list (g). Why?

is also because the producer "remembers the mode of production and is not executed immediately".
in generator g, the variable I always exists and has not been released and recycled, and then using the variable I to cycle the value of gpeni will cause confusion. In the first cycle, I is still 10, so the first value is 20. At this time, the I value has been assigned to 20, so when the second cycle executes iPlus, You get 41. In the same way, when you execute I + (iExj) for the third time, I is already 41, the fourth time. So the final output is 20pc41pyrus 84171

.

after changing the cyclic variable I to k, the variable name is no longer repeated and the assignment is no longer confused, which is consistent with the result of list (g)

.

this may be complicated to explain. The generator will execute the code only when it is needed, but another knowledge premise is needed here, that is, the variables in the deduction are temporary variables and will not affect other variables. Let's take a simple example: 

x = 5
a = (x for x in range(3))

print(list(a))    -sharp [0, 1, 2]
print(x)    -sharp 5

I don't remember which version of python dealt with this problem. There is no variable protection mechanism in some older versions, at least I used version 3.6. By the same token, let's take a look at this, which is closer to the title: 

g=(i for i in range(4))
i = 8
print(list(g))

guess what will be printed here, which is [0,1,2,3] . The I in the generator is protected, so it has nothing to do with the external variable I, and its value must be 0123.

look at another one: 

a = 1
g=(a + i for i in range(4))
a = 5
print(list(g))

here I is protected, but a is not. Since the value of subsequent an is 5, the generator in the print statement should be < gen 5 + 0,5 + 1,. > .

well, we can finally talk about the code in the title. g = (i for i in range (4) in the first line of
) , I is protected, so its iteration is always 0x3

. 
for i in [1,10]:
    g=(i+j for j in g)

the j here is also protected, and in the first loop, its value is the initial output of g. Here, I is not protected, but only binds variables to get its value when the generator generates the data.

value of g after the first loop: 

<gen i+0, i+1, i+2, i+3>

in the second cycle, j is the value of production. Although iTun10 at this time, I is subsequently bound, so the value of g after the second cycle of production:

. 
<gen i+i+0, i+i+1, i+i+2, i+i+3>

when the statement is finally printed, the value of I is the last value of the loop body (10), so the printout 20,21,22,23

by the way, upstairs: 

g=(i for i in range(4))
for i in [1,10]:
    g=(i+j for j in g)
for i in g:
    print(i)    -sharp20,41,84,171

give an explanation.

before the second loop body starts, the g generator is < gen iLoop 0, iLoop 1,., i+i+3 > .

for i in g indicates that the data generated from g is assigned to I. Get i+i+0 from the generator. To generate data, I have to substitute the value at this time. What is the value at this time? Yes, I is the last value (10) of the last loop body, so generate 20 and assign 20 to I;

the second loop fetches i+i+1 from the generator, where I is 20 (the last loop assigned the value, the difficulty is here), so the generation is 41, assign the value to I again;

third loop, get i+i+2 at this time the value of I is 41, so the resulting value is 84, which is assigned to I again;

. Wait

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