I have a list making up data, and I'd like to take 4 elements at a time from this list and put them in a 2d list where each 4-element increment is a new row of said list.
My first attempts involve input to 1d list:
list.append(input("Enter data type 1:")) list.append(input("Enter data type 2:")) etc.
and then I've tried to loop the list and to "switch" rows once the index reaches 4.
for x in range(n * 4):
for idx, y in enumerate(list):
if idx % 4 == 0:
x = x + 1
list[y] = result[x][y]
where I've initialised result according to the following:
and
ran = int(len(list)/4)
result=[[0 for x in range(ran)] for j in range(n)]
I've also attempted to ascribe a temporary empty list that will append to an initialised 2D list.
`
row.append(list)
result=[[x for x in row] for j in range(n + 1)]
#result[n]=row
print(result)
n = n + 1
row.clear()
list.clear()
so that each new loop starts with an empty row, takes input from user and copies it.
I'm at a loss for how to make result save the first entry and not be redefined at second,third,fourth entries.
I think this post is probably what you need. With np.reshape() you can just have your list filled with all the values you need and do the reshaping after in a single step.
Let's say I have an array:
print(arr1.shape)
(188621, 10)
And in the nth column (let's say 4 for this example), I want to check when a value is above a threshold, t. I want to create a new list (of x instances) of the entire row of arr1 when the ith iteration of the 4th column is above threshold t. In other words, it is extracting the ith row from arr1 when the condition in the 4th column is met. So far I have:
arr2 = []
for i in range(0,len(arr1)):
if arr1[i,4] > t:
arr2.append(arr1[i,:])
I have also tried something along the lines of:
for i in range(0,len(arr1)):
if arr1[i,4] > t:
if len(arr2) == 0:
arr2 = arr1[i,:]
else:
arr2 = np.concatenate((arr2,arr1[i,:]))
However, both instances seem to be growing in 1D terms of x*10 instead of a 2D list of (x, 10) when the conditions are met. What am I missing here?
Well, it wasn't that difficult.
arr2 = arr1[np.logical_not(arr1[:,4] < t)]
a=[2, 1, 3, 5, 3, 2]
def firstDuplicate(a):
for i in range(0,len(a)):
for j in range(i+1,len(a)):
while a[i]==a[j]:
num=[j]
break
print(num)
print(firstDuplicate(a))
The output should be coming as 4 and 5 but it's coming as 4 only
You can find the indices of all duplicates in an array in O(n) time and O(1) extra space with something like the following:
def get_duplicate_indices(arr):
inds = []
for i, val in enumerate(arr):
val = abs(val)
if arr[val] >= 0:
arr[val] = -arr[val]
else:
inds.append(i)
return inds
get_duplicate_indices(a)
[4, 5]
Note that this will modify the array in place! If you want to keep your input array un-modified, replace the first few lines in the above with:
def get_duplicate_indices(a):
arr = a.copy() # so we don't modify in place. Drawback is it's not O(n) extra space
inds = []
for i, val in enumerate(a):
# ...
Essentially this uses the sign of each element in the array as an indicator of whether a number has been seen before. If we come across a negative value, it means the number we reached has been seen before, so we append the number's index to our list of already-seen indices.
Note that this can run into trouble if the values in the array are larger than the length of the array, but in this case we just extend the working array to be the same length as whatever the maximum value is in the input. Easy peasy.
There are some things wrong with your code. The following will collect the indexes of every first duplicate:
def firstDuplicate(a):
num = [] # list to collect indexes of first dupes
for i in range(len(a)-1): # second to last is enough here
for j in range(i+1, len(a)):
if a[i]==a[j]: # while-loop made little sense
num.append(j) # grow list and do not override it
break # stop inner loop after first duplicate
print(num)
There are of course more performant algorithms to achieve this that are not quadratic.
I have an array of data where I'm taking a slice through a 3D array of floats and appending the numbers of the elements that satisfy a set of upper and lower bounds (also floats).
The first part of my code contains a nested for loop in the style of the following:
x_lst = []
for i in range(len(x1)):
for x in range(len(floatarray[0,:,0])):
if x1[i] <= floatarray[0,x,0] <= x2[i]:
x_lst.append(x)
#issues after this point
The inner loop compares the data in the array (floatarray[0,x,0]) with the boundaries x1 and x2, returning a list of integers, whilst the outer loop iterates through the values of the boundaries.
The issue I have is that the output (x_lst) is a single list, whereas I've been trying to produce a list of lists, with each sublist corresponding to the iterable i.
e.g.
#Desired output
x_lst = [[5,6,7,13,14],[21,22,23,36,37],[44,45,...
#Actual output
x_lst = [5,6,7,13,14,21,22,23,36,37,44,45,...
I feel like there's a very simple way of doing this, but I've not been able come up with anything that works (such as trying to use x_lst.split() after the append).
Also, any feedback on the question would be great as I'm still fairly new to SO.
It seems the following should work: why not create an intermediate sublist for each i, and add appropriate values to the sublist, then finally add the sublist to the main list?
x_lst = []
for i in range(len(x1)):
inner_list = [] # The sublist to be appended.
for x in range(len(floatarray[0,:,0])):
if x1[i] <= floatarray[0,x,0] <= x2[i]:
inner_list.append(x) # Add the numbers to the sublist.
# Issues after this point...
x_lst.append(inner_list) # Add the sublist to the main list.
Everything appears to be correct in the code except that you append into a 1-d array. For solving your problem you can simply change your code so that a temporary array named temp will append data in inner loop which would then be appended to your outer array x_lst as shown below:
x_lst = []
for i in range(len(x1))
temp=[]
for x in range(len(floatarray[0,:,0])):
if x1[i] <= floatarray[0,x,0] <= x2[i]:
temp.append(x)
x_lst.append(temp);
Okay, so Sam's answer is correct insofar as appending the values through the iteration, but trying it out with my code produced lists of lists of len(i) where each sublist contained all values of x that satisfied all elements of x1 and x2, instead of each list containing values specific to the [i]th element of x1 and x2, thus all the sublists were identical.
I got around this by replacing the inner for loop with a list comprehension, as follows, that gave the right result (in the desired form x_lst = [[5,6,7,13,14],[21,22,23,36,37],[44,45,...):
x_lst = []
for i in range(len(x1)):
y = [x for x in range(len(floatarray[0,:,0])) if x1[i] <= floatarray[0,x,0] <= x2[i]]
x_lst.append(y)
I am new to python. I think n begins at 0 and it will increase up till the last element on stack1.
arraylength = 3*stack1length
array = [0]*arraylength
for n in stack1:
array[3*n] = stack1[n]
my array length is 3 times the length of stack1
for n in stack1:
Goes through the items in stack1.
You seem to be wanting to go through the indexes:
for n in range(len(stack1)):
array[3*n] = stack1[n]
Note that this is better written with the convenience function, enumerate,
for n, stack1_n in enumerate(stack1):
array[3*n] = stack1_n
Additionally, you can use some evil hackz:
array[::3] = stack1
array[::3] is every third item in array (start:stop:step), and therefore you're setting every third item to the corresponding item in stack1.