Filter integers in numpy float array

up vote
18
down vote

favorite

Is there any built in function to discard integer and keep only float number in numpy.

import numpy as np

input = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

desired_ouput = some_function(input)
# Expected ouput
# desired_output = np.array([0.01, 2.001, 2.002])

asked Aug 30 at 10:14

has

503315

3

All values in that array are floats. "float" doesn't imply anything about the value being non-integer.
â€“Â user2357112
Aug 30 at 19:00

add a commentÂ |Â

up vote
18
down vote

favorite

Is there any built in function to discard integer and keep only float number in numpy.

import numpy as np

input = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

desired_ouput = some_function(input)
# Expected ouput
# desired_output = np.array([0.01, 2.001, 2.002])

asked Aug 30 at 10:14

has

503315

3

All values in that array are floats. "float" doesn't imply anything about the value being non-integer.
â€“Â user2357112
Aug 30 at 19:00

add a commentÂ |Â

up vote
18
down vote

favorite

Is there any built in function to discard integer and keep only float number in numpy.

import numpy as np

input = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

desired_ouput = some_function(input)
# Expected ouput
# desired_output = np.array([0.01, 2.001, 2.002])

asked Aug 30 at 10:14

has

503315

Is there any built in function to discard integer and keep only float number in numpy.

import numpy as np

input = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

desired_ouput = some_function(input)
# Expected ouput
# desired_output = np.array([0.01, 2.001, 2.002])

asked Aug 30 at 10:14

has

503315

asked Aug 30 at 10:14

has

503315

asked Aug 30 at 10:14

has

503315

asked Aug 30 at 10:14

has

503315

3

All values in that array are floats. "float" doesn't imply anything about the value being non-integer.
â€“Â user2357112
Aug 30 at 19:00

add a commentÂ |Â

3

All values in that array are floats. "float" doesn't imply anything about the value being non-integer.
â€“Â user2357112
Aug 30 at 19:00

All values in that array are floats. "float" doesn't imply anything about the value being non-integer.
â€“Â user2357112
Aug 30 at 19:00

add a commentÂ |Â

6 Answers
6

active

oldest

votes

up vote
13
down vote

accepted

Mask with whether each element is equal to it as an integer.

arr = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
out = arr[arr != arr.astype(int)]
#np.array([0.01, 2.001, 2.002])

answered Aug 30 at 10:23

Joe Iddon

13k31338

add a commentÂ |Â

up vote
12
down vote

I don't think so. My approach would be

import numpy as np
a = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
mask = np.isclose(a, a.astype(int))

print(a[~mask])
#[ 0.01 2.001 2.002]

answered Aug 30 at 10:21

SpghttCd

2,0372312

3

As a general rule always use isclose when comparing floats. I'm not sure it's necessary here, but upvoted for good practice.
â€“Â Daniel F
Aug 30 at 10:37

It's not a good practice to blindly follow what is called "good practice". isclose, depending on the source of these numbers, may actually be actively harmful.
â€“Â Ruslan
Aug 31 at 6:03

add a commentÂ |Â

up vote
5
down vote

I know of no in-built function. But you can create one yourself:

import numpy as np

A = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

def remove_ints(arr):
 return arr[~(arr == arr.astype(int))]

res = remove_ints(A)

array([ 0.01 , 2.001, 2.002])

Aside, you should not use a built-in class such as input as a variable name.

answered Aug 30 at 10:21

jpp

63.7k173780

add a commentÂ |Â

up vote
4
down vote

I've always used np.equal with np.mod:

>>> A[~np.equal(np.mod(A, 1), 0)]
array([0.01 , 2.001, 2.002])

edited Aug 30 at 10:55

answered Aug 30 at 10:40

user3483203

23.2k62147

add a commentÂ |Â

up vote
4
down vote

If you do not have to much data (short list), maybe do not need numpy:

>>> i = [0.0, 0.01, 1.0, 2.0, 2.001, 2.002]
>>> a=[j for j in i if not j.is_integer()]
>>> a
['0.01', '2.001', '2.002']

Otherwise see Joe Iddon answer

edited Sep 4 at 7:04

answered Aug 30 at 10:23

Dadep

2,24421631

10

The idea of scipy is to never use for for efficiency.
â€“Â Joe Iddon
Aug 30 at 10:25

add a commentÂ |Â

up vote
2
down vote

I don't know any builtin for this but you can filter those floats using:

filter(lambda x: int(str(x).split('.')[1]) != 0, input)

The lambda expression here checks if the decimal places are zero which I interpret as the number being an int.

edited Aug 30 at 10:29

MarianD

3,76861231

answered Aug 30 at 10:22

meissner_

36816

precision is lost by converting to string. e.g 2.001 becomes 2.0009999999999999.
â€“Â has
Aug 30 at 11:34

What? I just tried replicating this but [str(x) for x in input] returned: ['0.0', '0.01', '1.0', '2.0', '2.001', '2.002']? But even if i assume a loss of accuracy, it wouldn't change the results since '001' will not come out as 0 after the int cast, just like '00099999'.
â€“Â meissner_
Aug 30 at 11:42

It happend in my python. Actually these are indexes of pandas dataframe, so loss of precision mean, i can't get back values by apply .loc method.
â€“Â has
Aug 30 at 11:44

Well, this would be very interesting indeed if you can find a way to replicate the effect...
â€“Â meissner_
Aug 30 at 11:50

add a commentÂ |Â

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
13
down vote

accepted

Mask with whether each element is equal to it as an integer.

arr = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
out = arr[arr != arr.astype(int)]
#np.array([0.01, 2.001, 2.002])

answered Aug 30 at 10:23

Joe Iddon

13k31338

add a commentÂ |Â

up vote
13
down vote

accepted

Mask with whether each element is equal to it as an integer.

arr = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
out = arr[arr != arr.astype(int)]
#np.array([0.01, 2.001, 2.002])

answered Aug 30 at 10:23

Joe Iddon

13k31338

add a commentÂ |Â

up vote
13
down vote

accepted

Mask with whether each element is equal to it as an integer.

arr = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
out = arr[arr != arr.astype(int)]
#np.array([0.01, 2.001, 2.002])

answered Aug 30 at 10:23

Joe Iddon

13k31338

Mask with whether each element is equal to it as an integer.

arr = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
out = arr[arr != arr.astype(int)]
#np.array([0.01, 2.001, 2.002])

answered Aug 30 at 10:23

Joe Iddon

13k31338

answered Aug 30 at 10:23

Joe Iddon

13k31338

answered Aug 30 at 10:23

Joe Iddon

13k31338

answered Aug 30 at 10:23

Joe Iddon

13k31338

add a commentÂ |Â

up vote
12
down vote

I don't think so. My approach would be

import numpy as np
a = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
mask = np.isclose(a, a.astype(int))

print(a[~mask])
#[ 0.01 2.001 2.002]

answered Aug 30 at 10:21

SpghttCd

2,0372312

3

As a general rule always use isclose when comparing floats. I'm not sure it's necessary here, but upvoted for good practice.
â€“Â Daniel F
Aug 30 at 10:37

It's not a good practice to blindly follow what is called "good practice". isclose, depending on the source of these numbers, may actually be actively harmful.
â€“Â Ruslan
Aug 31 at 6:03

add a commentÂ |Â

up vote
12
down vote

I don't think so. My approach would be

import numpy as np
a = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
mask = np.isclose(a, a.astype(int))

print(a[~mask])
#[ 0.01 2.001 2.002]

answered Aug 30 at 10:21

SpghttCd

2,0372312

3

As a general rule always use isclose when comparing floats. I'm not sure it's necessary here, but upvoted for good practice.
â€“Â Daniel F
Aug 30 at 10:37

It's not a good practice to blindly follow what is called "good practice". isclose, depending on the source of these numbers, may actually be actively harmful.
â€“Â Ruslan
Aug 31 at 6:03

add a commentÂ |Â

up vote
12
down vote

I don't think so. My approach would be

import numpy as np
a = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
mask = np.isclose(a, a.astype(int))

print(a[~mask])
#[ 0.01 2.001 2.002]

answered Aug 30 at 10:21

SpghttCd

2,0372312

I don't think so. My approach would be

import numpy as np
a = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])
mask = np.isclose(a, a.astype(int))

print(a[~mask])
#[ 0.01 2.001 2.002]

answered Aug 30 at 10:21

SpghttCd

2,0372312

answered Aug 30 at 10:21

SpghttCd

2,0372312

answered Aug 30 at 10:21

SpghttCd

2,0372312

answered Aug 30 at 10:21

SpghttCd

2,0372312

3

As a general rule always use isclose when comparing floats. I'm not sure it's necessary here, but upvoted for good practice.
â€“Â Daniel F
Aug 30 at 10:37

It's not a good practice to blindly follow what is called "good practice". isclose, depending on the source of these numbers, may actually be actively harmful.
â€“Â Ruslan
Aug 31 at 6:03

add a commentÂ |Â

3

As a general rule always use isclose when comparing floats. I'm not sure it's necessary here, but upvoted for good practice.
â€“Â Daniel F
Aug 30 at 10:37

It's not a good practice to blindly follow what is called "good practice". isclose, depending on the source of these numbers, may actually be actively harmful.
â€“Â Ruslan
Aug 31 at 6:03

As a general rule always use isclose when comparing floats. I'm not sure it's necessary here, but upvoted for good practice.
â€“Â Daniel F
Aug 30 at 10:37

It's not a good practice to blindly follow what is called "good practice". isclose, depending on the source of these numbers, may actually be actively harmful.
â€“Â Ruslan
Aug 31 at 6:03

add a commentÂ |Â

up vote
5
down vote

I know of no in-built function. But you can create one yourself:

import numpy as np

A = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

def remove_ints(arr):
 return arr[~(arr == arr.astype(int))]

res = remove_ints(A)

array([ 0.01 , 2.001, 2.002])

Aside, you should not use a built-in class such as input as a variable name.

answered Aug 30 at 10:21

jpp

63.7k173780

add a commentÂ |Â

up vote
5
down vote

I know of no in-built function. But you can create one yourself:

import numpy as np

A = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

def remove_ints(arr):
 return arr[~(arr == arr.astype(int))]

res = remove_ints(A)

array([ 0.01 , 2.001, 2.002])

Aside, you should not use a built-in class such as input as a variable name.

answered Aug 30 at 10:21

jpp

63.7k173780

add a commentÂ |Â

up vote
5
down vote

I know of no in-built function. But you can create one yourself:

import numpy as np

A = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

def remove_ints(arr):
 return arr[~(arr == arr.astype(int))]

res = remove_ints(A)

array([ 0.01 , 2.001, 2.002])

Aside, you should not use a built-in class such as input as a variable name.

answered Aug 30 at 10:21

jpp

63.7k173780

I know of no in-built function. But you can create one yourself:

import numpy as np

A = np.array([0.0, 0.01, 1.0, 2.0, 2.001, 2.002])

def remove_ints(arr):
 return arr[~(arr == arr.astype(int))]

res = remove_ints(A)

array([ 0.01 , 2.001, 2.002])

Aside, you should not use a built-in class such as input as a variable name.

answered Aug 30 at 10:21

jpp

63.7k173780

answered Aug 30 at 10:21

jpp

63.7k173780

answered Aug 30 at 10:21

jpp

63.7k173780

answered Aug 30 at 10:21

jpp

63.7k173780

add a commentÂ |Â

up vote
4
down vote

I've always used np.equal with np.mod:

>>> A[~np.equal(np.mod(A, 1), 0)]
array([0.01 , 2.001, 2.002])

edited Aug 30 at 10:55

answered Aug 30 at 10:40

user3483203

23.2k62147

add a commentÂ |Â

up vote
4
down vote

I've always used np.equal with np.mod:

>>> A[~np.equal(np.mod(A, 1), 0)]
array([0.01 , 2.001, 2.002])

edited Aug 30 at 10:55

answered Aug 30 at 10:40

user3483203

23.2k62147

add a commentÂ |Â

up vote
4
down vote

I've always used np.equal with np.mod:

>>> A[~np.equal(np.mod(A, 1), 0)]
array([0.01 , 2.001, 2.002])

edited Aug 30 at 10:55

answered Aug 30 at 10:40

user3483203

23.2k62147

I've always used np.equal with np.mod:

>>> A[~np.equal(np.mod(A, 1), 0)]
array([0.01 , 2.001, 2.002])

edited Aug 30 at 10:55

answered Aug 30 at 10:40

user3483203

23.2k62147

edited Aug 30 at 10:55

answered Aug 30 at 10:40

user3483203

23.2k62147

answered Aug 30 at 10:40

user3483203

23.2k62147

answered Aug 30 at 10:40

user3483203

23.2k62147

add a commentÂ |Â

up vote
4
down vote

If you do not have to much data (short list), maybe do not need numpy:

>>> i = [0.0, 0.01, 1.0, 2.0, 2.001, 2.002]
>>> a=[j for j in i if not j.is_integer()]
>>> a
['0.01', '2.001', '2.002']

Otherwise see Joe Iddon answer

edited Sep 4 at 7:04

answered Aug 30 at 10:23

Dadep

2,24421631

10

The idea of scipy is to never use for for efficiency.
â€“Â Joe Iddon
Aug 30 at 10:25

add a commentÂ |Â

up vote
4
down vote

If you do not have to much data (short list), maybe do not need numpy:

>>> i = [0.0, 0.01, 1.0, 2.0, 2.001, 2.002]
>>> a=[j for j in i if not j.is_integer()]
>>> a
['0.01', '2.001', '2.002']

Otherwise see Joe Iddon answer

edited Sep 4 at 7:04

answered Aug 30 at 10:23

Dadep

2,24421631

10

The idea of scipy is to never use for for efficiency.
â€“Â Joe Iddon
Aug 30 at 10:25

add a commentÂ |Â

up vote
4
down vote

If you do not have to much data (short list), maybe do not need numpy:

>>> i = [0.0, 0.01, 1.0, 2.0, 2.001, 2.002]
>>> a=[j for j in i if not j.is_integer()]
>>> a
['0.01', '2.001', '2.002']

Otherwise see Joe Iddon answer

edited Sep 4 at 7:04

answered Aug 30 at 10:23

Dadep

2,24421631

If you do not have to much data (short list), maybe do not need numpy:

>>> i = [0.0, 0.01, 1.0, 2.0, 2.001, 2.002]
>>> a=[j for j in i if not j.is_integer()]
>>> a
['0.01', '2.001', '2.002']

Otherwise see Joe Iddon answer

edited Sep 4 at 7:04

answered Aug 30 at 10:23

Dadep

2,24421631

edited Sep 4 at 7:04

answered Aug 30 at 10:23

Dadep

2,24421631

answered Aug 30 at 10:23

Dadep

2,24421631

answered Aug 30 at 10:23

Dadep

2,24421631

10

The idea of scipy is to never use for for efficiency.
â€“Â Joe Iddon
Aug 30 at 10:25

add a commentÂ |Â

10

The idea of scipy is to never use for for efficiency.
â€“Â Joe Iddon
Aug 30 at 10:25

The idea of scipy is to never use for for efficiency.
â€“Â Joe Iddon
Aug 30 at 10:25

add a commentÂ |Â

up vote
2
down vote

I don't know any builtin for this but you can filter those floats using:

filter(lambda x: int(str(x).split('.')[1]) != 0, input)

The lambda expression here checks if the decimal places are zero which I interpret as the number being an int.

edited Aug 30 at 10:29

MarianD

3,76861231

answered Aug 30 at 10:22

meissner_

36816

precision is lost by converting to string. e.g 2.001 becomes 2.0009999999999999.
â€“Â has
Aug 30 at 11:34

What? I just tried replicating this but [str(x) for x in input] returned: ['0.0', '0.01', '1.0', '2.0', '2.001', '2.002']? But even if i assume a loss of accuracy, it wouldn't change the results since '001' will not come out as 0 after the int cast, just like '00099999'.
â€“Â meissner_
Aug 30 at 11:42

It happend in my python. Actually these are indexes of pandas dataframe, so loss of precision mean, i can't get back values by apply .loc method.
â€“Â has
Aug 30 at 11:44

Well, this would be very interesting indeed if you can find a way to replicate the effect...
â€“Â meissner_
Aug 30 at 11:50

add a commentÂ |Â

up vote
2
down vote

I don't know any builtin for this but you can filter those floats using:

filter(lambda x: int(str(x).split('.')[1]) != 0, input)

The lambda expression here checks if the decimal places are zero which I interpret as the number being an int.

edited Aug 30 at 10:29

MarianD

3,76861231

answered Aug 30 at 10:22

meissner_

36816

precision is lost by converting to string. e.g 2.001 becomes 2.0009999999999999.
â€“Â has
Aug 30 at 11:34

What? I just tried replicating this but [str(x) for x in input] returned: ['0.0', '0.01', '1.0', '2.0', '2.001', '2.002']? But even if i assume a loss of accuracy, it wouldn't change the results since '001' will not come out as 0 after the int cast, just like '00099999'.
â€“Â meissner_
Aug 30 at 11:42

It happend in my python. Actually these are indexes of pandas dataframe, so loss of precision mean, i can't get back values by apply .loc method.
â€“Â has
Aug 30 at 11:44

Well, this would be very interesting indeed if you can find a way to replicate the effect...
â€“Â meissner_
Aug 30 at 11:50

add a commentÂ |Â

up vote
2
down vote

I don't know any builtin for this but you can filter those floats using:

filter(lambda x: int(str(x).split('.')[1]) != 0, input)

The lambda expression here checks if the decimal places are zero which I interpret as the number being an int.

edited Aug 30 at 10:29

MarianD

3,76861231

answered Aug 30 at 10:22

meissner_

36816

I don't know any builtin for this but you can filter those floats using:

filter(lambda x: int(str(x).split('.')[1]) != 0, input)

The lambda expression here checks if the decimal places are zero which I interpret as the number being an int.

edited Aug 30 at 10:29

MarianD

3,76861231

answered Aug 30 at 10:22

meissner_

36816

edited Aug 30 at 10:29

MarianD

3,76861231

edited Aug 30 at 10:29

MarianD

3,76861231

edited Aug 30 at 10:29

MarianD

3,76861231

answered Aug 30 at 10:22

meissner_

36816

answered Aug 30 at 10:22

meissner_

36816

answered Aug 30 at 10:22

meissner_

36816

precision is lost by converting to string. e.g 2.001 becomes 2.0009999999999999.
â€“Â has
Aug 30 at 11:34

What? I just tried replicating this but [str(x) for x in input] returned: ['0.0', '0.01', '1.0', '2.0', '2.001', '2.002']? But even if i assume a loss of accuracy, it wouldn't change the results since '001' will not come out as 0 after the int cast, just like '00099999'.
â€“Â meissner_
Aug 30 at 11:42

It happend in my python. Actually these are indexes of pandas dataframe, so loss of precision mean, i can't get back values by apply .loc method.
â€“Â has
Aug 30 at 11:44

Well, this would be very interesting indeed if you can find a way to replicate the effect...
â€“Â meissner_
Aug 30 at 11:50

add a commentÂ |Â

precision is lost by converting to string. e.g 2.001 becomes 2.0009999999999999.
â€“Â has
Aug 30 at 11:34

What? I just tried replicating this but [str(x) for x in input] returned: ['0.0', '0.01', '1.0', '2.0', '2.001', '2.002']? But even if i assume a loss of accuracy, it wouldn't change the results since '001' will not come out as 0 after the int cast, just like '00099999'.
â€“Â meissner_
Aug 30 at 11:42

It happend in my python. Actually these are indexes of pandas dataframe, so loss of precision mean, i can't get back values by apply .loc method.
â€“Â has
Aug 30 at 11:44

Well, this would be very interesting indeed if you can find a way to replicate the effect...
â€“Â meissner_
Aug 30 at 11:50

precision is lost by converting to string. e.g 2.001 becomes 2.0009999999999999.
â€“Â has
Aug 30 at 11:34

What? I just tried replicating this but [str(x) for x in input] returned: ['0.0', '0.01', '1.0', '2.0', '2.001', '2.002']? But even if i assume a loss of accuracy, it wouldn't change the results since '001' will not come out as 0 after the int cast, just like '00099999'.
â€“Â meissner_
Aug 30 at 11:42

It happend in my python. Actually these are indexes of pandas dataframe, so loss of precision mean, i can't get back values by apply .loc method.
â€“Â has
Aug 30 at 11:44

Well, this would be very interesting indeed if you can find a way to replicate the effect...
â€“Â meissner_
Aug 30 at 11:50

add a commentÂ |Â

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