condition priority in if-else for efficiency
I would like to know if I have if-else or if-elif-elif-....-else condition like below:
if conditionA:
do A
elif conditionB:
do B
elif conditionC:
do C
...
...
...
else:
do z
Q1. If I already know that my condition resolve in conditionC for 99% of time, putting that condition as first one (instead of conditionA) would make my code more efficient?
Q2. Similarly, should I prioritize my statements that way, if Q1 is true?
Apologies if the question has already been asked. I might not have found the proper vocabulary to search for this.
Thank You.
python python-3.x performance if-statement big-o
asked Nov 23 '18 at 5:55
Jay Patel
463
add a comment |
I would like to know if I have if-else or if-elif-elif-....-else condition like below:
if conditionA:
do A
elif conditionB:
do B
elif conditionC:
do C
...
...
...
else:
do z
Q1. If I already know that my condition resolve in conditionC for 99% of time, putting that condition as first one (instead of conditionA) would make my code more efficient?
Q2. Similarly, should I prioritize my statements that way, if Q1 is true?
Apologies if the question has already been asked. I might not have found the proper vocabulary to search for this.
Thank You.
python python-3.x performance if-statement big-o
asked Nov 23 '18 at 5:55
Jay Patel
463
Have you tried testing this yourself?
– juanpa.arrivillaga
Nov 23 '18 at 5:56
Q1 - yes. Q2 - yes.
– b-fg
Nov 23 '18 at 5:57
add a comment |
I would like to know if I have if-else or if-elif-elif-....-else condition like below:
if conditionA:
do A
elif conditionB:
do B
elif conditionC:
do C
...
...
...
else:
do z
Q1. If I already know that my condition resolve in conditionC for 99% of time, putting that condition as first one (instead of conditionA) would make my code more efficient?
Q2. Similarly, should I prioritize my statements that way, if Q1 is true?
Apologies if the question has already been asked. I might not have found the proper vocabulary to search for this.
Thank You.
python python-3.x performance if-statement big-o
asked Nov 23 '18 at 5:55
Jay Patel
463
I would like to know if I have if-else or if-elif-elif-....-else condition like below:
if conditionA:
do A
elif conditionB:
do B
elif conditionC:
do C
...
...
...
else:
do z
Q1. If I already know that my condition resolve in conditionC for 99% of time, putting that condition as first one (instead of conditionA) would make my code more efficient?
Q2. Similarly, should I prioritize my statements that way, if Q1 is true?
Apologies if the question has already been asked. I might not have found the proper vocabulary to search for this.
Thank You.
python python-3.x performance if-statement big-o
python python-3.x performance if-statement big-o
asked Nov 23 '18 at 5:55
Jay Patel
463
asked Nov 23 '18 at 5:55
Jay Patel
463
asked Nov 23 '18 at 5:55
Jay Patel
463
asked Nov 23 '18 at 5:55
Jay Patel
463
asked Nov 23 '18 at 5:55
Jay Patel
463
463
Have you tried testing this yourself?
– juanpa.arrivillaga
Nov 23 '18 at 5:56
Q1 - yes. Q2 - yes.
– b-fg
Nov 23 '18 at 5:57
add a comment |
Have you tried testing this yourself?
– juanpa.arrivillaga
Nov 23 '18 at 5:56
Q1 - yes. Q2 - yes.
– b-fg
Nov 23 '18 at 5:57
Have you tried testing this yourself?
– juanpa.arrivillaga
Nov 23 '18 at 5:56
Have you tried testing this yourself?
– juanpa.arrivillaga
Nov 23 '18 at 5:56
Q1 - yes. Q2 - yes.
– b-fg
Nov 23 '18 at 5:57
Q1 - yes. Q2 - yes.
– b-fg
Nov 23 '18 at 5:57
add a comment |
2 Answers
2
active
oldest
votes
Q.1 and Q.2: Yes
According to Python documentation
An if … elif … elif … sequence is a substitute for the switch or case
statements found in other languages.
which means as soon as code executes a True condition, it will exit the if … elif … elif … sequence.
answered Nov 23 '18 at 6:33
ameydev
564
add a comment |
You see these conditions take O(1) time, you can say it is negligible to compared to the other code you have written along with it, so like if you have a code along with it which have two nested loops O(n^2) then the conditions are negligible time taking compared to your overall algo.
Moreover, this tells you the estimated complexity.
But other wise i'd say you must put it to condition A, because what you put in these conditions may have their respective time complexities.
suppose you have sub-string search using in operator in python,
it goes like.
st = 'hello' * (2 ** 999) # hellohellohellohe...
if 'hey' in st:
print('hey, I found it')
else:
print('well..')
it takes around θ(n) and O(MN), where m is string and n is sub-string to be compared.
have a look at this python substring.
So, it wrap this up, the preferred way would be to it on condition A.
answered Nov 23 '18 at 6:20
P.hunter
6171624
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Q.1 and Q.2: Yes
According to Python documentation
An if … elif … elif … sequence is a substitute for the switch or case
statements found in other languages.
which means as soon as code executes a True condition, it will exit the if … elif … elif … sequence.
answered Nov 23 '18 at 6:33
ameydev
564
add a comment |
Q.1 and Q.2: Yes
According to Python documentation
An if … elif … elif … sequence is a substitute for the switch or case
statements found in other languages.
which means as soon as code executes a True condition, it will exit the if … elif … elif … sequence.
answered Nov 23 '18 at 6:33
ameydev
564
add a comment |
Q.1 and Q.2: Yes
According to Python documentation
An if … elif … elif … sequence is a substitute for the switch or case
statements found in other languages.
which means as soon as code executes a True condition, it will exit the if … elif … elif … sequence.
answered Nov 23 '18 at 6:33
ameydev
564
Q.1 and Q.2: Yes
According to Python documentation
An if … elif … elif … sequence is a substitute for the switch or case
statements found in other languages.
which means as soon as code executes a True condition, it will exit the if … elif … elif … sequence.
answered Nov 23 '18 at 6:33
ameydev
564
answered Nov 23 '18 at 6:33
ameydev
564
answered Nov 23 '18 at 6:33
ameydev
564
answered Nov 23 '18 at 6:33
ameydev
564
564
add a comment |
add a comment |
You see these conditions take O(1) time, you can say it is negligible to compared to the other code you have written along with it, so like if you have a code along with it which have two nested loops O(n^2) then the conditions are negligible time taking compared to your overall algo.
Moreover, this tells you the estimated complexity.
But other wise i'd say you must put it to condition A, because what you put in these conditions may have their respective time complexities.
suppose you have sub-string search using in operator in python,
it goes like.
st = 'hello' * (2 ** 999) # hellohellohellohe...
if 'hey' in st:
print('hey, I found it')
else:
print('well..')
it takes around θ(n) and O(MN), where m is string and n is sub-string to be compared.
have a look at this python substring.
So, it wrap this up, the preferred way would be to it on condition A.
answered Nov 23 '18 at 6:20
P.hunter
6171624
add a comment |
You see these conditions take O(1) time, you can say it is negligible to compared to the other code you have written along with it, so like if you have a code along with it which have two nested loops O(n^2) then the conditions are negligible time taking compared to your overall algo.
Moreover, this tells you the estimated complexity.
But other wise i'd say you must put it to condition A, because what you put in these conditions may have their respective time complexities.
suppose you have sub-string search using in operator in python,
it goes like.
st = 'hello' * (2 ** 999) # hellohellohellohe...
if 'hey' in st:
print('hey, I found it')
else:
print('well..')
it takes around θ(n) and O(MN), where m is string and n is sub-string to be compared.
have a look at this python substring.
So, it wrap this up, the preferred way would be to it on condition A.
answered Nov 23 '18 at 6:20
P.hunter
6171624
add a comment |
You see these conditions take O(1) time, you can say it is negligible to compared to the other code you have written along with it, so like if you have a code along with it which have two nested loops O(n^2) then the conditions are negligible time taking compared to your overall algo.
Moreover, this tells you the estimated complexity.
But other wise i'd say you must put it to condition A, because what you put in these conditions may have their respective time complexities.
suppose you have sub-string search using in operator in python,
it goes like.
st = 'hello' * (2 ** 999) # hellohellohellohe...
if 'hey' in st:
print('hey, I found it')
else:
print('well..')
it takes around θ(n) and O(MN), where m is string and n is sub-string to be compared.
have a look at this python substring.
So, it wrap this up, the preferred way would be to it on condition A.
answered Nov 23 '18 at 6:20
P.hunter
6171624
You see these conditions take O(1) time, you can say it is negligible to compared to the other code you have written along with it, so like if you have a code along with it which have two nested loops O(n^2) then the conditions are negligible time taking compared to your overall algo.
Moreover, this tells you the estimated complexity.
But other wise i'd say you must put it to condition A, because what you put in these conditions may have their respective time complexities.
suppose you have sub-string search using in operator in python,
it goes like.
st = 'hello' * (2 ** 999) # hellohellohellohe...
if 'hey' in st:
print('hey, I found it')
else:
print('well..')
it takes around θ(n) and O(MN), where m is string and n is sub-string to be compared.
have a look at this python substring.
So, it wrap this up, the preferred way would be to it on condition A.
answered Nov 23 '18 at 6:20
P.hunter
6171624
answered Nov 23 '18 at 6:20
P.hunter
6171624
answered Nov 23 '18 at 6:20
P.hunter
6171624
answered Nov 23 '18 at 6:20
P.hunter
6171624
6171624
add a comment |
add a comment |
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Have you tried testing this yourself?
– juanpa.arrivillaga
Nov 23 '18 at 5:56
Q1 - yes. Q2 - yes.
– b-fg
Nov 23 '18 at 5:57