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how to specify range if we are taking an array in for loop

stackoverflow.com › questions › 48970416 › how-to-specify-range-if-we-are-taking-an-array-in-for-loop

There's various routes that you can take. Assuming you want to print the first 3 elements of your array, here's two:

array=[1,2,3,4,5]

1) Slice the array

# start at index 0 and go until index 2
for record in array[0:3]: 
    print(record)

2) Use a variable to iterate over the array

# range(3) = [0, 1, 2]
for i in range(3): 
    print(array[i])

Answer from randyjp on Stack Overflow

Stack Overflow

stackoverflow.com › questions › 48970416 › how-to-specify-range-if-we-are-taking-an-array-in-for-loop

python - how to specify range if we are taking an array in for loop - Stack Overflow

Top answer

1 of 2

4

There's various routes that you can take. Assuming you want to print the first 3 elements of your array, here's two:

array=[1,2,3,4,5]

1) Slice the array

# start at index 0 and go until index 2
for record in array[0:3]: 
    print(record)

2) Use a variable to iterate over the array

# range(3) = [0, 1, 2]
for i in range(3): 
    print(array[i])

2 of 2

0

In this code part-

array=[1,2,3,4,5]

for record in range(0,2,array): print(record)

The range function takes 3 parameters. First one is starting index,second one is ending index and the third one is the gap in which you want the numbers to be. So,going by an example-

for i in range(0,5,2): print(i)

This code will print 0,2,4 as answer. The range function will start from 0 and will return every 2nd number till 4(not 5,because 5 is excluded).

Your code gives an error because you have mentioned an array as your 3rd parameter but it is expecting an integer

for record in range(0,2,len(array)): print(record)

You could try this,but it will only return 0 because essentially this for loop simplifies to-

for record in range(0,2,5)

So after returning the starting number it will try to get the next 5th number which you dont have(as 2nd parameter is 2). Hope this clarifies.

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Using range() in accessing elements of an array

outcomes1000[range(0, i)] selects the first i elements of outcomes1000, though that's a very weird way of doing it. Typically, you'd just use slicing, so outcomes1000[:i]. Also, a better way to calculate the cumulative mean rather than with a list comprehension is with np.cumsum: averages = np.cumsum(outcomes1000) / np.arange(1, 1001) More on reddit.com

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[Python] How you would get a range when working with arrays.

r[1] is the 2nd element in the r tuple. In other words, it's the second dimension of x. In this case, the number 4: >>> x = numpy.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]]) >>> x 7: array([[ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12]]) >>> x.shape 8: (3L, 4L) so this line: for num in range(1,r[1]) is iterating through the columns of x More on reddit.com

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Top answer

1 of 5

3

outcomes1000[range(0, i)] selects the first i elements of outcomes1000, though that's a very weird way of doing it. Typically, you'd just use slicing, so outcomes1000[:i]. Also, a better way to calculate the cumulative mean rather than with a list comprehension is with np.cumsum: averages = np.cumsum(outcomes1000) / np.arange(1, 1001)

2 of 5

3

What does outcomes1000[range(0,i)] do? It returns a slice of the elements of outcomes1000 with the indexes between 0 and i-1. Can we use range() to access elements of an array? I mean, that's exactly what this code is doing, so clearly it's possible, but it seems unusual to use the range() call here instead of a more straightforward slice: averages = [np.mean(outcomes1000[:i]) for i in range(1,1001)]

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stackoverflow.com › questions › 12701659 › finding-range-of-a-numpy-array-elements

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Top answer

1 of 2

75

I think np.ptp might do what you want:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.ptp.html

Use with AIr = np.ptp(a,axis=1)

where r is your range array.

2 of 2

8

Try this:

Use with AIdef range_of_vals(x, axis=0):
    return np.max(x, axis=axis) - np.min(x, axis=axis)

reddit.com › r/learnprogramming › [python] how you would get a range when working with arrays.

r/learnprogramming on Reddit: [Python] How you would get a range when working with arrays.

January 26, 2018 -

I am taking a coding class with python for one of my grad school classes (Bio background). We are working in Spyder3

import numpy

x = numpy.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])

print(x[0,2])

r=x.shape
for num in range(1,r[1]):
    print(x[num-1,:])

print("Done!")

The above is the line of code my professor gave us in class. For the life of my I can't figure out where the (1,r[1]) comes from, I understand that r is referring back to what the variable "r" stands for but why are we using 1 to r[1] for the range? I know it stands for column and row of the matrix, the matrix is a 3,4. I hope if this is enough information, if not I will update!

Top answer

1 of 1

2

r[1] is the 2nd element in the r tuple. In other words, it's the second dimension of x. In this case, the number 4: >>> x = numpy.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]]) >>> x 7: array([[ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12]]) >>> x.shape 8: (3L, 4L) so this line: for num in range(1,r[1]) is iterating through the columns of x

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stackoverflow.com › questions › 18265935 › how-do-i-create-a-list-with-numbers-between-two-values

python - How do I create a list with numbers between two values? - Stack Overflow

Top answer

1 of 13

1045

Use range. In Python 2, it returns a list directly:

>>> range(11, 17)
[11, 12, 13, 14, 15, 16]

In Python 3, range is an iterator. To convert it to a list:

>>> list(range(11, 17))
[11, 12, 13, 14, 15, 16]

Note: The second number in range(start, stop) is exclusive. So, stop = 16+1 = 17.

To increment by steps of 0.5, consider using numpy's arange() and .tolist():

>>> import numpy as np
>>> np.arange(11, 17, 0.5).tolist()

[11.0, 11.5, 12.0, 12.5, 13.0, 13.5,
 14.0, 14.5, 15.0, 15.5, 16.0, 16.5]

See: How do I use a decimal step value for range()?

2 of 13

45

You seem to be looking for range():

>>> x1=11
>>> x2=16
>>> range(x1, x2+1)
[11, 12, 13, 14, 15, 16]
>>> list1 = range(x1, x2+1)
>>> list1
[11, 12, 13, 14, 15, 16]

For incrementing by 0.5 instead of 1, say:

>>> list2 = [x*0.5 for x in range(2*x1, 2*x2+1)]
>>> list2
[11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0]

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numpy.org › doc › stable › reference › generated › numpy.arange.html

numpy.arange — NumPy v2.5 Manual

>>> np.arange(0, 5, 0.5, dtype=np.int_) array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0]) >>> np.arange(-3, 3, 0.5, dtype=np.int_) array([-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8]) In such cases, the use of numpy.linspace should be preferred. The built-in range generates Python built-in integers that have arbitrary size, while numpy.arange produces numpy.int32 or numpy.int64 numbers.

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stackoverflow.com › questions › 13838081 › assigning-a-range-to-an-array-in-python

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Top answer

1 of 1

6

In Python 2.x:

arrayX = range(20080711, 20080714+1)

in Python 3.x:

arrayX = list(range(20080711, 20080714+1))

However, if your ints represent something like a date (YYYYMMDD), it will be trickier:

from datetime import datetime, timedelta
arrayX = []
dt = datetime(2008, 7, 11)
while dt <= datetime(2008, 7, 14):
    arrayX.append(int(dt.strftime('%Y%m%d')))
    dt += timedelta(days=1)

which works over months and years.

GeeksforGeeks

geeksforgeeks.org › python › numpy-arrange-in-python

numpy.arange() in Python - GeeksforGeeks

February 14, 2026 - numpy.arange() function creates an array of evenly spaced values within a given interval. It is similar to Python's built-in range() function but returns a NumPy array instead of a list.