Using a variable before it’s been defined
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Why does the code below work when pop_zero and pop_end use the list variable, x, before x has even been defined (x = list(range(i))). In other words pop_zero and pop_end are defined in terms of x, but x is defined after pop_zero and pop_end. Why isn’t this a problem?
import timeit
from timeit import Timer
pop_zero = Timer("x.pop(0)", "from __main__ import x")
pop_end = Timer("x.pop()", "from __main__ import x")
print("pop(0) pop()")
for i in range(1000000,100000001,1000000):
x = list(range(i))
pt = pop_end.timeit(number=1000)
x = list(range(i))
pz = pop_zero.timeit(number=1000)
print("%15.5f, %15.5f" %(pz,pt))
python
asked Nov 9 at 21:32
user3124200
674
add a comment |
up vote
2
down vote
favorite
Why does the code below work when pop_zero and pop_end use the list variable, x, before x has even been defined (x = list(range(i))). In other words pop_zero and pop_end are defined in terms of x, but x is defined after pop_zero and pop_end. Why isn’t this a problem?
import timeit
from timeit import Timer
pop_zero = Timer("x.pop(0)", "from __main__ import x")
pop_end = Timer("x.pop()", "from __main__ import x")
print("pop(0) pop()")
for i in range(1000000,100000001,1000000):
x = list(range(i))
pt = pop_end.timeit(number=1000)
x = list(range(i))
pz = pop_zero.timeit(number=1000)
print("%15.5f, %15.5f" %(pz,pt))
python
asked Nov 9 at 21:32
user3124200
674
3
bothpop_ ...functions usexas a string and the string doesn't get evaluated until you invoke the function
– JacobIRR
Nov 9 at 21:33
@JacobIRR either it's a dupe or you should flesh out an answer IMO because I'm confused on this
– roganjosh
Nov 9 at 21:37
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Why does the code below work when pop_zero and pop_end use the list variable, x, before x has even been defined (x = list(range(i))). In other words pop_zero and pop_end are defined in terms of x, but x is defined after pop_zero and pop_end. Why isn’t this a problem?
import timeit
from timeit import Timer
pop_zero = Timer("x.pop(0)", "from __main__ import x")
pop_end = Timer("x.pop()", "from __main__ import x")
print("pop(0) pop()")
for i in range(1000000,100000001,1000000):
x = list(range(i))
pt = pop_end.timeit(number=1000)
x = list(range(i))
pz = pop_zero.timeit(number=1000)
print("%15.5f, %15.5f" %(pz,pt))
python
asked Nov 9 at 21:32
user3124200
674
Why does the code below work when pop_zero and pop_end use the list variable, x, before x has even been defined (x = list(range(i))). In other words pop_zero and pop_end are defined in terms of x, but x is defined after pop_zero and pop_end. Why isn’t this a problem?
import timeit
from timeit import Timer
pop_zero = Timer("x.pop(0)", "from __main__ import x")
pop_end = Timer("x.pop()", "from __main__ import x")
print("pop(0) pop()")
for i in range(1000000,100000001,1000000):
x = list(range(i))
pt = pop_end.timeit(number=1000)
x = list(range(i))
pz = pop_zero.timeit(number=1000)
print("%15.5f, %15.5f" %(pz,pt))
python
python
asked Nov 9 at 21:32
user3124200
674
asked Nov 9 at 21:32
user3124200
674
asked Nov 9 at 21:32
user3124200
674
asked Nov 9 at 21:32
user3124200
674
asked Nov 9 at 21:32
user3124200
674
674
3
bothpop_ ...functions usexas a string and the string doesn't get evaluated until you invoke the function
– JacobIRR
Nov 9 at 21:33
@JacobIRR either it's a dupe or you should flesh out an answer IMO because I'm confused on this
– roganjosh
Nov 9 at 21:37
add a comment |
3
bothpop_ ...functions usexas a string and the string doesn't get evaluated until you invoke the function
– JacobIRR
Nov 9 at 21:33
@JacobIRR either it's a dupe or you should flesh out an answer IMO because I'm confused on this
– roganjosh
Nov 9 at 21:37
3
3
both
pop_ ...functions use x as a string and the string doesn't get evaluated until you invoke the function– JacobIRR
Nov 9 at 21:33
both
pop_ ...functions use x as a string and the string doesn't get evaluated until you invoke the function– JacobIRR
Nov 9 at 21:33
@JacobIRR either it's a dupe or you should flesh out an answer IMO because I'm confused on this
– roganjosh
Nov 9 at 21:37
@JacobIRR either it's a dupe or you should flesh out an answer IMO because I'm confused on this
– roganjosh
Nov 9 at 21:37
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
When you create a Timer object, it doesn't actually execute the code inside yet. Only when you call one of its method (i.e. .timeit()) does it actually run the code.
answered Nov 9 at 21:37
Tomothy32
4878
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
When you create a Timer object, it doesn't actually execute the code inside yet. Only when you call one of its method (i.e. .timeit()) does it actually run the code.
answered Nov 9 at 21:37
Tomothy32
4878
add a comment |
up vote
1
down vote
When you create a Timer object, it doesn't actually execute the code inside yet. Only when you call one of its method (i.e. .timeit()) does it actually run the code.
answered Nov 9 at 21:37
Tomothy32
4878
add a comment |
up vote
1
down vote
up vote
1
down vote
When you create a Timer object, it doesn't actually execute the code inside yet. Only when you call one of its method (i.e. .timeit()) does it actually run the code.
answered Nov 9 at 21:37
Tomothy32
4878
When you create a Timer object, it doesn't actually execute the code inside yet. Only when you call one of its method (i.e. .timeit()) does it actually run the code.
answered Nov 9 at 21:37
Tomothy32
4878
answered Nov 9 at 21:37
Tomothy32
4878
answered Nov 9 at 21:37
Tomothy32
4878
answered Nov 9 at 21:37
Tomothy32
4878
4878
add a comment |
add a comment |
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pop_ ...functions usexas a string and the string doesn't get evaluated until you invoke the function– JacobIRR
Nov 9 at 21:33
@JacobIRR either it's a dupe or you should flesh out an answer IMO because I'm confused on this
– roganjosh
Nov 9 at 21:37