Related
List if i have two lists l1 = [ 'A', 'B' ]
l2 = [ 1, 2 ]
What is the most elegant way to get a pandas dataframe like this: +-----+-----+-----+
| | l1 | l2 |
+-----+-----+-----+
| 0 | A | 1 |
+-----+-----+-----+
| 1 | A | 2 |
+-----+-----+--
GhostKU I'm not sure my question is correct, but I don't know how to explain it. So I have something like a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
And I need to get something like: result = [
['11', '21', '31'],
['11', '21', '32'],
['11'
srsrso I have two lists: ints = [10, 20, 30, 40, 50]
and opers = ['+', '-', '*', '/']
I would like to get a list containing all possible combinations of these two lists, for example: 10+20*40-30/50 = 810
50-40+30*20/10 = 70
etc The list should be[810, 70,
wei Given I have two lists: val ints = listOf(0, 1, 2)
val strings = listOf("a", "b", "c")
I want all possible combinations of their elements 0a, 1a, 2a, 0betc Is there a more elegant way than: ints.forEach { int ->
strings.forEach { string ->
Bryant I have a list of 40,000 elements, of which only 16 are unique, but all weighted differently. letters = letters[1:16]
freq=c(8221, 6717, 4912, 3810, 3409, 2807, 2607, 2406, 2206, 1203, 501, 401, 301, 201, 201, 100)
list = rep(letters,fr
Daniel Rollins I have a list of elements (I have 11 elements in my actual list, this is just an example): x <- c(1, 2, 3)
and want to split them into two lists (with all the entries), but I want to return all possible combinations of this list like: (1,2)(3)
Ravi I'm trying to get all combinations of a list where two elements that are next to each other could "be one". E.g; >>> list = ['a', 'b', 'c', 'd']
>>> get_comb(list)
[['ab', 'c' 'd'], ['a', 'bc' 'd'], ['a', 'b', 'cd'] ['ab', 'cd']]
The tricky part is that
GhostKU I'm not sure my question is correct, but I don't know how to explain it. So I have something like a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
And I need to get something like: result = [
['11', '21', '31'],
['11', '21', '32'],
['11'
GhostKU I'm not sure my question is correct, but I don't know how to explain it. So I have something like a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
And I need to get something like: result = [
['11', '21', '31'],
['11', '21', '32'],
['11'
wei Given I have two lists: val ints = listOf(0, 1, 2)
val strings = listOf("a", "b", "c")
I want all possible combinations of their elements 0a, 1a, 2a, 0betc Is there a more elegant way than: ints.forEach { int ->
strings.forEach { string ->
Peaceful Kegan Liberation I'm really sorry for the title, but I don't know how to describe my problem in words. So here is the example: Assuming we have a string "123", my function should yield: 1 2 3
12 3
1 23
or for the string "1234": 1 2 3 4
12 3 4
12 34
1
username I'm pretty sure there's an easy answer to this, but due to my limited R experience I'm having a hard time solving it. I have a list of dataframes representing different experiments, and for each of these dataframes I generate a regression model - the
Bryant I have a list of 40,000 elements, of which only 16 are unique, but all weighted differently. letters = letters[1:16]
freq=c(8221, 6717, 4912, 3810, 3409, 2807, 2607, 2406, 2206, 1203, 501, 401, 301, 201, 201, 100)
list = rep(letters,fr
Ravi I'm trying to get all combinations of a list where two elements that are next to each other could "be one". E.g; >>> list = ['a', 'b', 'c', 'd']
>>> get_comb(list)
[['ab', 'c' 'd'], ['a', 'bc' 'd'], ['a', 'b', 'cd'] ['ab', 'cd']]
The tricky part is that
Fear Robert I have a question about python. I have a grid like the attached image. I have coordinates for all states. The coordinates of the states are shown in the A and B lists: A = [1,2,3]
B = [4,5,6]
I want to find the coordinates of the state I want. For
List if i have two lists l1 = [ 'A', 'B' ]
l2 = [ 1, 2 ]
What is the most elegant way to get a pandas dataframe like this: +-----+-----+-----+
| | l1 | l2 |
+-----+-----+-----+
| 0 | A | 1 |
+-----+-----+-----+
| 1 | A | 2 |
+-----+-----+--
Bryant I have a list of 40,000 elements, of which only 16 are unique, but all weighted differently. letters = letters[1:16]
freq=c(8221, 6717, 4912, 3810, 3409, 2807, 2607, 2406, 2206, 1203, 501, 401, 301, 201, 201, 100)
list = rep(letters,fr
Ravi I'm trying to get all combinations of a list where two elements that are next to each other could "be one". E.g; >>> list = ['a', 'b', 'c', 'd']
>>> get_comb(list)
[['ab', 'c' 'd'], ['a', 'bc' 'd'], ['a', 'b', 'cd'] ['ab', 'cd']]
The tricky part is that
List if i have two lists l1 = [ 'A', 'B' ]
l2 = [ 1, 2 ]
What is the most elegant way to get a pandas dataframe like this: +-----+-----+-----+
| | l1 | l2 |
+-----+-----+-----+
| 0 | A | 1 |
+-----+-----+-----+
| 1 | A | 2 |
+-----+-----+--
GhostKU I'm not sure my question is correct, but I don't know how to explain it. So I have something like a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
And I need to get something like: result = [
['11', '21', '31'],
['11', '21', '32'],
['11'
wei Given I have two lists: val ints = listOf(0, 1, 2)
val strings = listOf("a", "b", "c")
I want all possible combinations of their elements 0a, 1a, 2a, 0band many more Is there a more elegant way than: ints.forEach { int ->
strings.forEach { string ->
wei Given I have two lists: val ints = listOf(0, 1, 2)
val strings = listOf("a", "b", "c")
I want all possible combinations of their elements 0a, 1a, 2a, 0band many more Is there a more elegant way than: ints.forEach { int ->
strings.forEach { string ->
Daniel Rollins I have a list of elements (I have 11 elements in my actual list, this is just an example): x <- c(1, 2, 3)
and want to split them into two lists (with all entries), but I want to return all possible combinations of this list like: (1,2)(3) & (1
Bryant I have a list of 40,000 elements, of which only 16 are unique, but all weighted differently. letters = letters[1:16]
freq=c(8221, 6717, 4912, 3810, 3409, 2807, 2607, 2406, 2206, 1203, 501, 401, 301, 201, 201, 100)
list = rep(letters,fr
Bryant I have a list of 40,000 elements, of which only 16 are unique, but all weighted differently. letters = letters[1:16]
freq=c(8221, 6717, 4912, 3810, 3409, 2807, 2607, 2406, 2206, 1203, 501, 401, 301, 201, 201, 100)
list = rep(letters,fr
Peaceful Kegan Liberation I'm really sorry for the title, but I don't know how to describe my problem in words. So here is the example: Assuming we have a string "123", my function should yield: 1 2 3
12 3
1 23
or for the string "1234": 1 2 3 4
12 3 4
12 34
1
Ravi I'm trying to get all combinations of a list where two elements that are next to each other could "be one". E.g; >>> list = ['a', 'b', 'c', 'd']
>>> get_comb(list)
[['ab', 'c' 'd'], ['a', 'bc' 'd'], ['a', 'b', 'cd'] ['ab', 'cd']]
The tricky part is that
username I'm pretty sure there's an easy answer to this, but due to my limited R experience I'm having a hard time solving it. I have a list of dataframes representing different experiments, and for each of these dataframes I generate a regression model - the
Ravi I'm trying to get all combinations of a list where two elements that are next to each other could "be one". E.g; >>> list = ['a', 'b', 'c', 'd']
>>> get_comb(list)
[['ab', 'c' 'd'], ['a', 'bc' 'd'], ['a', 'b', 'cd'] ['ab', 'cd']]
The tricky part is that