Find number of times a number is followed by a larger number in a list
I have this function to find the number of times a number is followed by a larger number in a list. Is there another more "pythonic" way this could be done? I am using Python 3.7.0.
Thanks in advance.
def find_greater_numbers(arr):
count = 0
i = 0
j = 1
while i < len(arr):
while j < len(arr):
if arr[j] > arr[i]:
count += 1
j+=1
j = i+1
i+=1
return count
find_greater_numbers([6,1,2,7]]) # returns 4
python python-3.7
edited Nov 13 '18 at 3:58
Julien
7,60331537
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
add a comment |
I have this function to find the number of times a number is followed by a larger number in a list. Is there another more "pythonic" way this could be done? I am using Python 3.7.0.
Thanks in advance.
def find_greater_numbers(arr):
count = 0
i = 0
j = 1
while i < len(arr):
while j < len(arr):
if arr[j] > arr[i]:
count += 1
j+=1
j = i+1
i+=1
return count
find_greater_numbers([6,1,2,7]]) # returns 4
python python-3.7
edited Nov 13 '18 at 3:58
Julien
7,60331537
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
add a comment |
I have this function to find the number of times a number is followed by a larger number in a list. Is there another more "pythonic" way this could be done? I am using Python 3.7.0.
Thanks in advance.
def find_greater_numbers(arr):
count = 0
i = 0
j = 1
while i < len(arr):
while j < len(arr):
if arr[j] > arr[i]:
count += 1
j+=1
j = i+1
i+=1
return count
find_greater_numbers([6,1,2,7]]) # returns 4
python python-3.7
edited Nov 13 '18 at 3:58
Julien
7,60331537
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
I have this function to find the number of times a number is followed by a larger number in a list. Is there another more "pythonic" way this could be done? I am using Python 3.7.0.
Thanks in advance.
def find_greater_numbers(arr):
count = 0
i = 0
j = 1
while i < len(arr):
while j < len(arr):
if arr[j] > arr[i]:
count += 1
j+=1
j = i+1
i+=1
return count
find_greater_numbers([6,1,2,7]]) # returns 4
python python-3.7
python python-3.7
edited Nov 13 '18 at 3:58
Julien
7,60331537
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
edited Nov 13 '18 at 3:58
Julien
7,60331537
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
edited Nov 13 '18 at 3:58
Julien
7,60331537
edited Nov 13 '18 at 3:58
Julien
7,60331537
edited Nov 13 '18 at 3:58
Julien
7,60331537
7,60331537
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
asked Nov 13 '18 at 3:43
johnsmithoptionaljohnsmithoptional
82
82
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It's a bit unclear if you mean immediately followed, or followed at any later index
In the first case, this one liner:
sum(x < y for x,y in zip(arr[:-1],arr[1:])) # answer is 2
In the second, this one:
sum(any(x < y for y in arr[i:]) for i,x in enumerate(arr)) # answer is 3
And if you want to count the number of exact such pairs (like what your actual code seems to be doing):
sum(x < y for i,x in enumerate(arr) for y in arr[i:]) # answer is 4
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
add a comment |
Use map to get the list of True/False for each pair of number in the list depending on the condition if x < y. This list is being generated using the for loop. After getting the list, filter the number of 'True' in the list. Then create a single list using itertools.chain.from_iterable() function and find its length.
import itertools
arr = [6, 1, 2, 7]
num = len(list(itertools.chain.from_iterable(list(filter(lambda x: x , map(lambda x, y: x<y, arr[:-i], arr[i:]))) for i, x in enumerate(arr))))
print(num)
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
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
It's a bit unclear if you mean immediately followed, or followed at any later index
In the first case, this one liner:
sum(x < y for x,y in zip(arr[:-1],arr[1:])) # answer is 2
In the second, this one:
sum(any(x < y for y in arr[i:]) for i,x in enumerate(arr)) # answer is 3
And if you want to count the number of exact such pairs (like what your actual code seems to be doing):
sum(x < y for i,x in enumerate(arr) for y in arr[i:]) # answer is 4
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
add a comment |
It's a bit unclear if you mean immediately followed, or followed at any later index
In the first case, this one liner:
sum(x < y for x,y in zip(arr[:-1],arr[1:])) # answer is 2
In the second, this one:
sum(any(x < y for y in arr[i:]) for i,x in enumerate(arr)) # answer is 3
And if you want to count the number of exact such pairs (like what your actual code seems to be doing):
sum(x < y for i,x in enumerate(arr) for y in arr[i:]) # answer is 4
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
add a comment |
It's a bit unclear if you mean immediately followed, or followed at any later index
In the first case, this one liner:
sum(x < y for x,y in zip(arr[:-1],arr[1:])) # answer is 2
In the second, this one:
sum(any(x < y for y in arr[i:]) for i,x in enumerate(arr)) # answer is 3
And if you want to count the number of exact such pairs (like what your actual code seems to be doing):
sum(x < y for i,x in enumerate(arr) for y in arr[i:]) # answer is 4
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
It's a bit unclear if you mean immediately followed, or followed at any later index
In the first case, this one liner:
sum(x < y for x,y in zip(arr[:-1],arr[1:])) # answer is 2
In the second, this one:
sum(any(x < y for y in arr[i:]) for i,x in enumerate(arr)) # answer is 3
And if you want to count the number of exact such pairs (like what your actual code seems to be doing):
sum(x < y for i,x in enumerate(arr) for y in arr[i:]) # answer is 4
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
edited Nov 13 '18 at 4:07
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
answered Nov 13 '18 at 3:53
JulienJulien
7,60331537
7,60331537
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
add a comment |
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
Hello. I meant followed by a greater number at any later index. So in the example list [6,1,2,7], 6 is followed 1 time by a greater number 1 is followed 2 times by a greater number , 2 is followed 1 time by a greater number (1+2+1=4). Thanks.
– johnsmithoptional
Nov 13 '18 at 4:24
add a comment |
Use map to get the list of True/False for each pair of number in the list depending on the condition if x < y. This list is being generated using the for loop. After getting the list, filter the number of 'True' in the list. Then create a single list using itertools.chain.from_iterable() function and find its length.
import itertools
arr = [6, 1, 2, 7]
num = len(list(itertools.chain.from_iterable(list(filter(lambda x: x , map(lambda x, y: x<y, arr[:-i], arr[i:]))) for i, x in enumerate(arr))))
print(num)
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
add a comment |
Use map to get the list of True/False for each pair of number in the list depending on the condition if x < y. This list is being generated using the for loop. After getting the list, filter the number of 'True' in the list. Then create a single list using itertools.chain.from_iterable() function and find its length.
import itertools
arr = [6, 1, 2, 7]
num = len(list(itertools.chain.from_iterable(list(filter(lambda x: x , map(lambda x, y: x<y, arr[:-i], arr[i:]))) for i, x in enumerate(arr))))
print(num)
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
add a comment |
Use map to get the list of True/False for each pair of number in the list depending on the condition if x < y. This list is being generated using the for loop. After getting the list, filter the number of 'True' in the list. Then create a single list using itertools.chain.from_iterable() function and find its length.
import itertools
arr = [6, 1, 2, 7]
num = len(list(itertools.chain.from_iterable(list(filter(lambda x: x , map(lambda x, y: x<y, arr[:-i], arr[i:]))) for i, x in enumerate(arr))))
print(num)
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
Use map to get the list of True/False for each pair of number in the list depending on the condition if x < y. This list is being generated using the for loop. After getting the list, filter the number of 'True' in the list. Then create a single list using itertools.chain.from_iterable() function and find its length.
import itertools
arr = [6, 1, 2, 7]
num = len(list(itertools.chain.from_iterable(list(filter(lambda x: x , map(lambda x, y: x<y, arr[:-i], arr[i:]))) for i, x in enumerate(arr))))
print(num)
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
answered Nov 13 '18 at 4:33
Rishabh MishraRishabh Mishra
378310
378310
add a comment |
add a comment |
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