Keep updating list while summing

-3

I'm a begginer in Python. I need help.
I recieve a starting point (in this case, (2, 0)) and must sum 2 numbers to that. Eg (2, 0) + (1, 1) = (3, 1).
I have to keep doing this till I achieve a maximum value (in this case 5) of elements is the list.

But I have a problem. I can't make a loop and keep updating the list.
The function must work with any value. Can someone help me?
Here is what I have (know it's wrong, but i'm stuck):

def summ(a, b):

    return (a[0] + b[0], a[1] + b[1])



x= (((5), (2,0), (1,1)))



maxx= x[0] #max of elements in list

start= x[1] #starting point

direction= x[2] #what it must sum



def position(x):



    new_list= ()



    for i in x:

        new_list = start, summ(start, direction)

        new_list += (summ(new_list[-1], direction), )

        if len(new_list) == maxx:

            break



    return new_list



#the output I had

((2, 0), (3, 1), (4, 2))



#the output I need

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4)) #5 elements, maxx

asked Jan 19 at 2:57

Kelly Baptista

1

move the new_list = start, summ(start, direction) outside of the for loop.

– Stephen Rauch
Jan 19 at 3:03

add a comment |

-3

But I have a problem. I can't make a loop and keep updating the list.
The function must work with any value. Can someone help me?
Here is what I have (know it's wrong, but i'm stuck):

def summ(a, b):

    return (a[0] + b[0], a[1] + b[1])



x= (((5), (2,0), (1,1)))



maxx= x[0] #max of elements in list

start= x[1] #starting point

direction= x[2] #what it must sum



def position(x):



    new_list= ()



    for i in x:

        new_list = start, summ(start, direction)

        new_list += (summ(new_list[-1], direction), )

        if len(new_list) == maxx:

            break



    return new_list



#the output I had

((2, 0), (3, 1), (4, 2))



#the output I need

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4)) #5 elements, maxx

asked Jan 19 at 2:57

Kelly Baptista

1

move the new_list = start, summ(start, direction) outside of the for loop.

– Stephen Rauch
Jan 19 at 3:03

add a comment |

-3

But I have a problem. I can't make a loop and keep updating the list.
The function must work with any value. Can someone help me?
Here is what I have (know it's wrong, but i'm stuck):

def summ(a, b):

    return (a[0] + b[0], a[1] + b[1])



x= (((5), (2,0), (1,1)))



maxx= x[0] #max of elements in list

start= x[1] #starting point

direction= x[2] #what it must sum



def position(x):



    new_list= ()



    for i in x:

        new_list = start, summ(start, direction)

        new_list += (summ(new_list[-1], direction), )

        if len(new_list) == maxx:

            break



    return new_list



#the output I had

((2, 0), (3, 1), (4, 2))



#the output I need

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4)) #5 elements, maxx

asked Jan 19 at 2:57

Kelly Baptista

But I have a problem. I can't make a loop and keep updating the list.
The function must work with any value. Can someone help me?
Here is what I have (know it's wrong, but i'm stuck):

def summ(a, b):

    return (a[0] + b[0], a[1] + b[1])



x= (((5), (2,0), (1,1)))



maxx= x[0] #max of elements in list

start= x[1] #starting point

direction= x[2] #what it must sum



def position(x):



    new_list= ()



    for i in x:

        new_list = start, summ(start, direction)

        new_list += (summ(new_list[-1], direction), )

        if len(new_list) == maxx:

            break



    return new_list



#the output I had

((2, 0), (3, 1), (4, 2))



#the output I need

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4)) #5 elements, maxx

python

asked Jan 19 at 2:57

Kelly Baptista

asked Jan 19 at 2:57

Kelly Baptista

asked Jan 19 at 2:57

Kelly Baptista

asked Jan 19 at 2:57

Kelly Baptista

asked Jan 19 at 2:57

Kelly Baptista

1

move the new_list = start, summ(start, direction) outside of the for loop.

– Stephen Rauch
Jan 19 at 3:03

add a comment |

1

move the new_list = start, summ(start, direction) outside of the for loop.

– Stephen Rauch
Jan 19 at 3:03

move the new_list = start, summ(start, direction) outside of the for loop.

– Stephen Rauch
Jan 19 at 3:03

add a comment |

3 Answers
3

active

oldest

votes

-1

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]





x = 5, (2, 0), (1, 1)





def position(x):

    i = 1

    new_list = [x[i]]



    while i < x[0]:

        new_list += summ(new_list[i - 1], x[2]),

        i = i + 1



    return new_list





print(position(x))

Tested and the OUTPUT is

[(2, 0), (3, 1), (4, 2), (5, 3), (6, 4)]

edited Jan 19 at 4:15

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

1

I just added to the return: tuple(new_list). So the output will be ((2,0), (3,1), (4,2), (5,3), (6,4)). Thank you! :)

– Kelly Baptista
Jan 19 at 19:50

add a comment |

Use a list comprehension:

a = ((5,), (2,0), (1,1))



b = [(a[1][0] + i*a[2][0], a[1][1] + i*a[2][1]) for i in range(a[0][0])]

edited Jan 19 at 10:21

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

add a comment |

-1

Let's take @StephenRauch's sage advice and move the:

new_list = start, summ(start, direction)

outside of the for loop. And clean up the code as we do:

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]



def position(argument):



    maxx, start, direction = argument



    new_list = [start]



    for _ in range(maxx - len(new_list)):

        new_list.append(summ(new_list[-1], direction))



    return tuple(new_list)



print(position(((5), (2, 0), (1, 1))))

OUTPUT

> python3 test.py

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4))

>

Now let's have some more fun with the code and rather than reinvent vector arithmetic, we'll import Vec2D from Python turtle. And return a list which makes more sense to me than a tuple:

from turtle import Vec2D



def position(argument):



    maxx, start, direction = argument



    return [start + n * direction for n in range(maxx)]



print(position(((5), Vec2D(2, 0), Vec2D(1, 1))))

OUTPUT

> python3 test2.py

[(2.00,0.00), (3.00,1.00), (4.00,2.00), (5.00,3.00), (6.00,4.00)]

>

edited Jan 20 at 22:20

answered Jan 19 at 6:04

cdlane

17.9k21144

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

-1

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]





x = 5, (2, 0), (1, 1)





def position(x):

    i = 1

    new_list = [x[i]]



    while i < x[0]:

        new_list += summ(new_list[i - 1], x[2]),

        i = i + 1



    return new_list





print(position(x))

Tested and the OUTPUT is

[(2, 0), (3, 1), (4, 2), (5, 3), (6, 4)]

edited Jan 19 at 4:15

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

1

I just added to the return: tuple(new_list). So the output will be ((2,0), (3,1), (4,2), (5,3), (6,4)). Thank you! :)

– Kelly Baptista
Jan 19 at 19:50

add a comment |

-1

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]





x = 5, (2, 0), (1, 1)





def position(x):

    i = 1

    new_list = [x[i]]



    while i < x[0]:

        new_list += summ(new_list[i - 1], x[2]),

        i = i + 1



    return new_list





print(position(x))

Tested and the OUTPUT is

[(2, 0), (3, 1), (4, 2), (5, 3), (6, 4)]

edited Jan 19 at 4:15

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

1

I just added to the return: tuple(new_list). So the output will be ((2,0), (3,1), (4,2), (5,3), (6,4)). Thank you! :)

– Kelly Baptista
Jan 19 at 19:50

add a comment |

-1

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]





x = 5, (2, 0), (1, 1)





def position(x):

    i = 1

    new_list = [x[i]]



    while i < x[0]:

        new_list += summ(new_list[i - 1], x[2]),

        i = i + 1



    return new_list





print(position(x))

Tested and the OUTPUT is

[(2, 0), (3, 1), (4, 2), (5, 3), (6, 4)]

edited Jan 19 at 4:15

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]





x = 5, (2, 0), (1, 1)





def position(x):

    i = 1

    new_list = [x[i]]



    while i < x[0]:

        new_list += summ(new_list[i - 1], x[2]),

        i = i + 1



    return new_list





print(position(x))

Tested and the OUTPUT is

[(2, 0), (3, 1), (4, 2), (5, 3), (6, 4)]

edited Jan 19 at 4:15

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

edited Jan 19 at 4:15

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

answered Jan 19 at 3:36

Hassan Voyeau

1,69021720

1

I just added to the return: tuple(new_list). So the output will be ((2,0), (3,1), (4,2), (5,3), (6,4)). Thank you! :)

– Kelly Baptista
Jan 19 at 19:50

add a comment |

1

I just added to the return: tuple(new_list). So the output will be ((2,0), (3,1), (4,2), (5,3), (6,4)). Thank you! :)

– Kelly Baptista
Jan 19 at 19:50

I just added to the return: tuple(new_list). So the output will be ((2,0), (3,1), (4,2), (5,3), (6,4)). Thank you! :)

– Kelly Baptista
Jan 19 at 19:50

add a comment |

Use a list comprehension:

a = ((5,), (2,0), (1,1))



b = [(a[1][0] + i*a[2][0], a[1][1] + i*a[2][1]) for i in range(a[0][0])]

edited Jan 19 at 10:21

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

add a comment |

Use a list comprehension:

a = ((5,), (2,0), (1,1))



b = [(a[1][0] + i*a[2][0], a[1][1] + i*a[2][1]) for i in range(a[0][0])]

edited Jan 19 at 10:21

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

add a comment |

Use a list comprehension:

a = ((5,), (2,0), (1,1))



b = [(a[1][0] + i*a[2][0], a[1][1] + i*a[2][1]) for i in range(a[0][0])]

edited Jan 19 at 10:21

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

Use a list comprehension:

a = ((5,), (2,0), (1,1))



b = [(a[1][0] + i*a[2][0], a[1][1] + i*a[2][1]) for i in range(a[0][0])]

edited Jan 19 at 10:21

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

edited Jan 19 at 10:21

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

answered Jan 19 at 9:43

Jan Christoph Terasa

1,9501022

add a comment |

-1

Let's take @StephenRauch's sage advice and move the:

new_list = start, summ(start, direction)

outside of the for loop. And clean up the code as we do:

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]



def position(argument):



    maxx, start, direction = argument



    new_list = [start]



    for _ in range(maxx - len(new_list)):

        new_list.append(summ(new_list[-1], direction))



    return tuple(new_list)



print(position(((5), (2, 0), (1, 1))))

OUTPUT

> python3 test.py

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4))

>

Now let's have some more fun with the code and rather than reinvent vector arithmetic, we'll import Vec2D from Python turtle. And return a list which makes more sense to me than a tuple:

from turtle import Vec2D



def position(argument):



    maxx, start, direction = argument



    return [start + n * direction for n in range(maxx)]



print(position(((5), Vec2D(2, 0), Vec2D(1, 1))))

OUTPUT

> python3 test2.py

[(2.00,0.00), (3.00,1.00), (4.00,2.00), (5.00,3.00), (6.00,4.00)]

>

edited Jan 20 at 22:20

answered Jan 19 at 6:04

cdlane

17.9k21144

add a comment |

-1

Let's take @StephenRauch's sage advice and move the:

new_list = start, summ(start, direction)

outside of the for loop. And clean up the code as we do:

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]



def position(argument):



    maxx, start, direction = argument



    new_list = [start]



    for _ in range(maxx - len(new_list)):

        new_list.append(summ(new_list[-1], direction))



    return tuple(new_list)



print(position(((5), (2, 0), (1, 1))))

OUTPUT

> python3 test.py

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4))

>

Now let's have some more fun with the code and rather than reinvent vector arithmetic, we'll import Vec2D from Python turtle. And return a list which makes more sense to me than a tuple:

from turtle import Vec2D



def position(argument):



    maxx, start, direction = argument



    return [start + n * direction for n in range(maxx)]



print(position(((5), Vec2D(2, 0), Vec2D(1, 1))))

OUTPUT

> python3 test2.py

[(2.00,0.00), (3.00,1.00), (4.00,2.00), (5.00,3.00), (6.00,4.00)]

>

edited Jan 20 at 22:20

answered Jan 19 at 6:04

cdlane

17.9k21144

add a comment |

-1

Let's take @StephenRauch's sage advice and move the:

new_list = start, summ(start, direction)

outside of the for loop. And clean up the code as we do:

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]



def position(argument):



    maxx, start, direction = argument



    new_list = [start]



    for _ in range(maxx - len(new_list)):

        new_list.append(summ(new_list[-1], direction))



    return tuple(new_list)



print(position(((5), (2, 0), (1, 1))))

OUTPUT

> python3 test.py

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4))

>

Now let's have some more fun with the code and rather than reinvent vector arithmetic, we'll import Vec2D from Python turtle. And return a list which makes more sense to me than a tuple:

from turtle import Vec2D



def position(argument):



    maxx, start, direction = argument



    return [start + n * direction for n in range(maxx)]



print(position(((5), Vec2D(2, 0), Vec2D(1, 1))))

OUTPUT

> python3 test2.py

[(2.00,0.00), (3.00,1.00), (4.00,2.00), (5.00,3.00), (6.00,4.00)]

>

edited Jan 20 at 22:20

answered Jan 19 at 6:04

cdlane

17.9k21144

Let's take @StephenRauch's sage advice and move the:

new_list = start, summ(start, direction)

outside of the for loop. And clean up the code as we do:

def summ(a, b):

    return a[0] + b[0], a[1] + b[1]



def position(argument):



    maxx, start, direction = argument



    new_list = [start]



    for _ in range(maxx - len(new_list)):

        new_list.append(summ(new_list[-1], direction))



    return tuple(new_list)



print(position(((5), (2, 0), (1, 1))))

OUTPUT

> python3 test.py

((2, 0), (3, 1), (4, 2), (5, 3), (6, 4))

>

Now let's have some more fun with the code and rather than reinvent vector arithmetic, we'll import Vec2D from Python turtle. And return a list which makes more sense to me than a tuple:

from turtle import Vec2D



def position(argument):



    maxx, start, direction = argument



    return [start + n * direction for n in range(maxx)]



print(position(((5), Vec2D(2, 0), Vec2D(1, 1))))

OUTPUT

> python3 test2.py

[(2.00,0.00), (3.00,1.00), (4.00,2.00), (5.00,3.00), (6.00,4.00)]

>

edited Jan 20 at 22:20

answered Jan 19 at 6:04

cdlane

17.9k21144

edited Jan 20 at 22:20

answered Jan 19 at 6:04

cdlane

17.9k21144

answered Jan 19 at 6:04

cdlane

17.9k21144

answered Jan 19 at 6:04

cdlane

17.9k21144

add a comment |

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