selecting array elements of certain range and define a new array

Question

I am not very used to pyhton and want to write a function that takes an array x as an input and gives an array back (select) consisting only of those entries of the input array that fulfil a certain property, e.g. being in a certain range. The function that should do this is the following:

def select(x):
    count = 0
    select = []                #0
    for i in range(0,len(x[0])):
      if ( (int(x[4][i])+0.5) > x[4][i] > (int(x[4][i])-0.5)  ):
        select[0][count]=x[0][i] #1
        select[1][count]=x[1][i] #2
        select[2][count]=x[4][i] #3
        count = count + 1
    return select

However, if I call that function I get the following error message:

IndexError: list index out of range

the line causing it is "#1" (and the 2 following lines are making trouble too I think). I guess I have to define the array size in some way. How can I do that in python in that case? As I see select=[] is not enough.

Kind regards

The issue (as I think you realize) is select has no [0] element. Can you show what you expect the input x and output select to be for some case? — Brian
– Brian, Commented Jan 13, 2016 at 14:03

jatinderjit · Accepted Answer · 2016-01-13 14:18:49Z

1

select is an empty list initially. You are trying trying to assign values to its elements that don't currently exist.

Probably you need select = [[], [], []]

And elements of inner elements of the select won't also exist when you try to assign them in #1, #2 and #3

Perhaps this is what you want:

def select(x):
    select = [[] for i in range(3)]    #0 : [[], [], []]
    for i in range(0,len(x[0])):
      if ( (int(x[4][i])+0.5) > x[4][i] > (int(x[4][i])-0.5)  ):
        select[0].append(x[0][i]) #1
        select[1].append(x[1][i]) #2
        select[2].append(x[4][i]) #3
    return select

edited Jan 13, 2016 at 14:18

answered Jan 13, 2016 at 14:04

jatinderjit

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4 Comments

Brian Over a year ago

Perhaps, but seems like an odd choice to blank fill it with. I think select = range(3) is the more pythonic way.

jatinderjit Over a year ago

He wants to nest something within select, so a dict or another list is required, not a range.

Brian Over a year ago

Totally, gotcha, duh!

DonkeyKong Over a year ago

@jatinderjit Thanks a lot!

dfrib · Accepted Answer · 2016-01-13 14:07:55Z

1

As an alternative to your own filter method (to achieve array filtering w.r.t. a range), you can use the following approach

myRange = [8, 11]
myArr = [6, 7, 8, 9, 10, 11, 12]
myArrFiltered = [x for x in myArr if myRange[0] <= x <= myRange[1]]
print(myArrFiltered)
# [8, 9, 10, 11]

answered Jan 13, 2016 at 14:07

dfrib

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Comments

masnun · Accepted Answer · 2016-01-13 14:10:51Z

1

Your code does not work because select list has no items at index 0.

If I were to implement similar functionality I would have used filter with a function that decides which elements should be selected:

ages = [5, 15, 3, 7, 18, 34, 9, 10]

def check_age(age):
    # we could implement all sorts of logic here
    # if we return True, the element is selected
    return age > 10

children_over_ten = filter(check_age, ages)

print(children_over_ten)

Or simply use list comprehension if the selection criteria is simple enough:

ages = [5, 15, 3, 7, 18, 34, 9, 10]
children_over_ten = [x for x in ages if x > 10]
print(children_over_ten)

answered Jan 13, 2016 at 14:10

masnun

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Comments

J.J. Hakala · Accepted Answer · 2016-01-13 14:13:00Z

1

You should probably try using list comprehensions, which can be used for filtering data.

For example with interactive python session,

>>> data = [i*0.25 for i in range(0, 21)]
>>> data
[0.0, 0.25, 0.5, 0.75, 1.0, 1.25, 1.5, 1.75, 2.0, 2.25, 2.5, 2.75, 3.0, 3.25, 3.5, 3.75, 4.0, 4.25, 4.5, 4.75, 5.0]
>>> [x for x in data if abs(2.5 - x) <= 0.5]
[2.0, 2.25, 2.5, 2.75, 3.0]

which selects the data that is in range [2.0, 3.0].

Iterating through an array can be often accomplished without using indexes, for example

for x in array:
    print(x)

answered Jan 13, 2016 at 14:13

J.J. Hakala

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Comments

Quinn · Accepted Answer · 2016-01-13 14:30:04Z

1

Or declare select = [], and do it this way:

def select(x):
    count = 0
    select = [] #0
    for i in range(0,len(x[0])):
      if ( (int(x[4][i])+0.5) > x[4][i] > (int(x[4][i])-0.5)  ):
        row = [x[r*r][i] for r in range(3)]
        select.append(row)
        count = count + 1
    return select

answered Jan 13, 2016 at 14:30

Quinn

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Comments

innoSPG · Accepted Answer · 2016-01-13 14:56:02Z

1

You need to initialize select accordingly

def select(x):
    count = 0
    select = [[],[],[]]                #0
    for i in range(0,len(x[0])):
      if ( (int(x[4][i])+0.5) > x[4][i] > (int(x[4][i])-0.5)  ):
        select[0]+=[x[0][i]] #1
        select[1]+=[x[1][i]] #2
        select[2]+=[x[4][i]] #3
        count = count + 1
    return select

edited Jan 13, 2016 at 14:56

answered Jan 13, 2016 at 14:16

innoSPG

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3 Comments

DonkeyKong Over a year ago

thanks for your time, following this I get TypeError: You must first set_array for mappable, any ideas? To be honest I don't know what += does.

innoSPG Over a year ago

a += b is a shorthan for a = a+b. For list, + does the concatenation.

innoSPG Over a year ago

@DonkeyKong, I just edited the question I did not make the right hand side a list. I just added [] arround the rhs. You can also use append instead of the concatenation.

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