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Return indices when list stops increasing Python
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Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
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I have been trying to return the indices where my lists(arrays) stop increasing. So far I can only get the index of the first time the list stops increasing and it returns it repeatedly. With array_1 = np.array([1,2,2,1,1,2,1]) the output should be [2,5] since those are the indices that stop increasing.
def monotonic_check(ori_array):
indices_array = []
for i in ori_array:
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
This code instead is returning [2,2,2]
python python-3.x python-2.x
asked Mar 8 at 23:00
KenKen
93
add a comment |
I have been trying to return the indices where my lists(arrays) stop increasing. So far I can only get the index of the first time the list stops increasing and it returns it repeatedly. With array_1 = np.array([1,2,2,1,1,2,1]) the output should be [2,5] since those are the indices that stop increasing.
def monotonic_check(ori_array):
indices_array = []
for i in ori_array:
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
This code instead is returning [2,2,2]
python python-3.x python-2.x
asked Mar 8 at 23:00
KenKen
93
For i in range(l'en(array))
– BlueSheepToken
Mar 8 at 23:05
add a comment |
I have been trying to return the indices where my lists(arrays) stop increasing. So far I can only get the index of the first time the list stops increasing and it returns it repeatedly. With array_1 = np.array([1,2,2,1,1,2,1]) the output should be [2,5] since those are the indices that stop increasing.
def monotonic_check(ori_array):
indices_array = []
for i in ori_array:
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
This code instead is returning [2,2,2]
python python-3.x python-2.x
asked Mar 8 at 23:00
KenKen
93
I have been trying to return the indices where my lists(arrays) stop increasing. So far I can only get the index of the first time the list stops increasing and it returns it repeatedly. With array_1 = np.array([1,2,2,1,1,2,1]) the output should be [2,5] since those are the indices that stop increasing.
def monotonic_check(ori_array):
indices_array = []
for i in ori_array:
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
This code instead is returning [2,2,2]
python python-3.x python-2.x
python python-3.x python-2.x
asked Mar 8 at 23:00
KenKen
93
asked Mar 8 at 23:00
KenKen
93
asked Mar 8 at 23:00
KenKen
93
asked Mar 8 at 23:00
KenKen
93
asked Mar 8 at 23:00
KenKen
93
93
For i in range(l'en(array))
– BlueSheepToken
Mar 8 at 23:05
add a comment |
For i in range(l'en(array))
– BlueSheepToken
Mar 8 at 23:05
For i in range(l'en(array))
– BlueSheepToken
Mar 8 at 23:05
For i in range(l'en(array))
– BlueSheepToken
Mar 8 at 23:05
add a comment |
1 Answer
1
active
oldest
votes
You are using the value instead of the index
def monotonic_check(ori_array):
indices_array = []
for i in range(len(ori_array)):
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are using the value instead of the index
def monotonic_check(ori_array):
indices_array = []
for i in range(len(ori_array)):
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
add a comment |
You are using the value instead of the index
def monotonic_check(ori_array):
indices_array = []
for i in range(len(ori_array)):
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
add a comment |
You are using the value instead of the index
def monotonic_check(ori_array):
indices_array = []
for i in range(len(ori_array)):
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
You are using the value instead of the index
def monotonic_check(ori_array):
indices_array = []
for i in range(len(ori_array)):
try:
if ori_array[i] > ori_array[i+1]:
indices_array.append(i)
else:
continue
except:
pass
return indices_array
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
answered Mar 8 at 23:03
Pablo MartinezPablo Martinez
3215
3215
add a comment |
add a comment |
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For i in range(l'en(array))
– BlueSheepToken
Mar 8 at 23:05