Is there a functional way to do this?
>
I have an array [0,1,2,3,0,1,2,2,3] and I want the first chunk of
non-decreasing values from this array (eg: In this case I want
[0,1,2,3])
Sounds like a use for a generator wrapper:
def monotonic(iterator):
i = iter(iterator)
prev = i.next()
yield
prev
while True:
this = i.next()
if prev <= this:
yield this
prev = this
else:
break
>>lst1 = [0,1,2,3,0,1,2,2,3]
lst2 = [0,1,2,3,3,0,1,2,2,3]
print list(monotonic(lst1))
[0,1,2,3]
>>print list(monotonic(lst2))
[0,1,2,3,3]
Adjust the "<=" to "<" depending on your needs
(my "lst2"
condition probes that edge)
-tkc
Oct 8 '08 #1
I am trying to split a list of numbers into sublists once a condition is met.
num_list = [0,1,2,3,4,5,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5]
Whenever the list reaches 5, it needs to be splitted as a sublist resulting as below:
[[0,1,2,3,4,5],[2,3,4,5],[0,1,2,3,4,5],[0,1,2,3,4,5]]
I tried the code below which seems to work but it places 5 into the following list and I can't figure out how to place it in the previous list instead.
num_list =[0,1,2,3,4,5,1,2,3,4,5,2,3,4,5] arrays = [[num_list[0]]] # array of sub-arrays (starts with first value) for i in range(1, len(num_list)): # go through each element after the first if num_list[i] != 5: # If it's larger than the previous arrays[len(arrays)-1].append(num_list[i]) # Add it to the last sub-array else: # otherwise arrays.append([num_list[i]]) # Make a new sub-array print(arrays)used from the solution given in this link: //stackoverflow.com/questions/5255...-condition
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num_list = [0,1,2,3,4,5,1,2,3,4,5,2,3,4,5] arrays = [[]] # array of sub-arrays for i, num in enumerate(num_list): # go through each element after the first arrays[-1].append(num) # Add it to the last sub-array if num == 5 and i != len(num_list)-1: # if 5 encountered and not last element arrays.append([]) print(arrays)
Output:
Output:
[array([0, 1, 2, 3, 4, 5]), array([1, 2, 3, 4, 5]), array([2, 3, 4, 5])]Posts: 53
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May-25-2019, 05:45 PM (This post was last modified: May-25-2019, 05:48 PM by python_newbie09.)
you are very helpful! thank you very much.
(May-25-2019, 11:27 AM)michalmonday Wrote: num_list = [0,1,2,3,4,5,1,2,3,4,5,2,3,4,5] arrays = [[]] # array of sub-arrays for i, num in enumerate(num_list): # go through each element after the first arrays[-1].append(num) # Add it to the last sub-array if num == 5 and i != len(num_list)-1: # if 5 encountered and not last element arrays.append([]) print(arrays)Output:
[[0, 1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [2, 3, 4, 5]]import numpy as np num_arr = np.array([0,1,2,3,4,5,1,2,3,4,5,2,3,4,5]) arrays = np.split(num_arr, np.where(num_arr[:-1] == 5)[0]+1) print(arrays)Output:
[array([0, 1, 2, 3, 4, 5]), array([1, 2, 3, 4, 5]), array([2, 3, 4, 5])]
btw, could explain what is actually happening in this line of code, especially the last part [0]+1
arrays = np.split(num_arr, np.where(num_arr[:-1] == 5)[0]+1)Posts: 4
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Thank you for the help.I am trying to get extra credit on an assignment and this was very helpful.
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I hope this clarifies what's going on in there, but the best way is to check every part of the expression yourself by printing parts of it (or also modifying/playing with them) using interpreter
import numpy as np # np.array allows to use "array-oriented" math operations # For example using: # num_arr == 5 # Doing that to a list will return single "True" or "False" # Doing that to np.array will return another array containing "True" or "False" # corresponding to each value in the array # [5,10] == 5 evaluates to "False" # np.array([5,10]) == 5 evaluates to "array([True, False])" # So if you want to easily apply some operation (possibly using some numpy # functions) to an array then using np.array() makes it convenient. num_arr = np.array([0,1,2,3,4,5,1,2,3,4,5,2,3,4,5]) indices_where_5_is_found = np.where(num_arr[:-1] == 5)[0] + 1 print('Array:', num_arr, '\n') print('At which indices 5 was found:\n', indices_where_5_is_found, '\n') # Notice that if we didn't use [:-1] then the index of the last 5 # would also be included (resulting in additional empty sub-list), # [:-1] removes the last item from the array. print("At which indices 5 would be found if [:-1] wasn't used:\n", np.where(num_arr == 5)[0] + 1, '\n') print("Result if [:-1] wasn't used:\n", np.split(num_arr, np.where(num_arr == 5)[0] + 1), '\n') # If +1 wasn't used then 5 would be included as the first item of each # sub-list # np.array([1,2]) + 1 increases each item of the array by 1, resulting in: # array([2,3]) print("Indices if +1 wasn't used:\n", np.where(num_arr[:-1] == 5)[0], '\n') print("Result if +1 wasn't used:\n", np.split(num_arr, np.where(num_arr[:-1] == 5)[0]), '\n') arrays = np.split(num_arr, indices_where_5_is_found) print('Final result:\n', arrays, '\n')Output:
Array: [0 1 2 3 4 5 1 2 3 4 5 2 3 4 5] At which indices 5 was found: [ 6 11] At which indices 5 would be found if [:-1] wasn't used: [ 6 11 15] Result if [:-1] wasn't used: [array([0, 1, 2, 3, 4, 5]), array([1, 2, 3, 4, 5]), array([2, 3, 4, 5]), array([], dtype=int32)] Indices if +1 wasn't used: [ 5 10] Result if +1 wasn't used: [array([0, 1, 2, 3, 4]), array([5, 1, 2, 3, 4]), array([5, 2, 3, 4, 5])] Final result: [array([0, 1, 2, 3, 4, 5]), array([1, 2, 3, 4, 5]), array([2, 3, 4, 5])]Posts: 53
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May-26-2019, 06:42 PM (This post was last modified: May-26-2019, 07:36 PM by python_newbie09.)
thanks a lot for the explanation and the tip to print out the results. will try it as well
I would like to extend this problem to another level where once these sublists are created. I need to iterate through this sublist and extract the index whenever there is a break in sequence. I tried to do that myself but I am not able to come up with a good solution for it. I know it will be easier to create a separate function and to pass the results of this separated arrays to is but beyond that I am not sure how best to proceed. As an example I have the array as below:
array = [0,1,2,3,4,5,6,7,8,9,10,0,1,2,3,4,5,6,7,3,8,9,10]
and with your help, I achieved the following:
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 3, 8, 9, 10]]
Now I would like to loop into this sublists and when it find a break in the sequence within the sub list it will extract the index and value of it. in this case the break of sequence occurs in the second sublist from 7 to 3. I tried something like below which resulted in TypeError: append() takes exactly one argument (0 given)
The expected output in this case is 3
arrays = [[]] # array of sub-arrays for i, num in enumerate(array): arrays[-1].append(num) if num == 10 and i != len(array)-1: arrays.append([]) print(arrays) seq_break = [] for i in arrays: if i[-1] != i[-1] + 1: #checks to see if the next value in the sublist equals to the previous value +1 which means a sequence, if not extract the break value seq_break.append() print(seq_break)Posts: 95
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You could do it like this:
array = [0,1,2,3,4,5,6,7,8,9,10,0,1,2,3,4,5,6,7,3,8,9,10] arrays = [[]] # array of sub-arrays for i, num in enumerate(array): arrays[-1].append(num) if num == 10 and i != len(array)-1: arrays.append([]) print(arrays) seq_break_list = [] for array in arrays: seq_break = [] for i, value in enumerate(array): if i == 0: continue if value != array[i-1] + 1: #checks to see if the next value in the sublist equals to the previous value +1 which means a sequence, if not extract the break value seq_break.append((i, value)) seq_break_list.append(seq_break) print(seq_break_list)Output:
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [0, 1, 2, 3, 4, 5, 6, 7, 3, 8, 9, 10]] [[], [(8, 3), (9, 8)]]Posts: 53
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Joined: Jan 2018
Reputation: 4
thanks i tried this approach but i realise that my data is much more complicated than initially observed.
I have situations whereby the sub arrays would look like below:
[[0,1,2,3,4,5,6,7,8,9,10],[0,1,2,3,4,5,6,6,7,7,8,8,9,10][0,1,2,3,4,5,6,7,8,9][0,1,2,3,4,5,6,7,1,8,9,10]
So there are situations where the number repeats itself or the ending number could be either 9 or 10 which makes the earlier code to split the array not work if the ending number is 9. and what i am trying to extract from each of this sequence is when there is an obvious break in it like going from 7 to 1 in the last subarray example. I am not sure how best to deal with this now given the complexity of these sub arrays.