Looping through a linked list containing arrays in python










1















i'm struggling to work out how to use a loop to run through a list which contains 4 array values being "a_value", "b_value", "c_value", "d_value". Each of my arrays in this list have 23 elements. the outcome im looking for is for a loop to run though the list adding all 23 elements of each array. E.G. total of "a_value" then total of "b_value" and so on. this is what i have so far.
Thank you guys.



class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None

class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value) # each '""_value' has 23 integer values
in them
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

arraylist.listprint()

i = 1
x = 1
total_a = 0
total_b = 0
total_c = 0
total_d = 0

for n in arraylist:
for z in range(len(n)-1)
value_a = int(a_value[i])
total_a+=value_a
i+= 1
x+=1









share|improve this question
























  • What is your input and expected output?

    – Alex
    Nov 14 '18 at 14:15











  • Did it help you?

    – NoorJafri
    Nov 14 '18 at 14:36















1















i'm struggling to work out how to use a loop to run through a list which contains 4 array values being "a_value", "b_value", "c_value", "d_value". Each of my arrays in this list have 23 elements. the outcome im looking for is for a loop to run though the list adding all 23 elements of each array. E.G. total of "a_value" then total of "b_value" and so on. this is what i have so far.
Thank you guys.



class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None

class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value) # each '""_value' has 23 integer values
in them
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

arraylist.listprint()

i = 1
x = 1
total_a = 0
total_b = 0
total_c = 0
total_d = 0

for n in arraylist:
for z in range(len(n)-1)
value_a = int(a_value[i])
total_a+=value_a
i+= 1
x+=1









share|improve this question
























  • What is your input and expected output?

    – Alex
    Nov 14 '18 at 14:15











  • Did it help you?

    – NoorJafri
    Nov 14 '18 at 14:36













1












1








1








i'm struggling to work out how to use a loop to run through a list which contains 4 array values being "a_value", "b_value", "c_value", "d_value". Each of my arrays in this list have 23 elements. the outcome im looking for is for a loop to run though the list adding all 23 elements of each array. E.G. total of "a_value" then total of "b_value" and so on. this is what i have so far.
Thank you guys.



class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None

class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value) # each '""_value' has 23 integer values
in them
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

arraylist.listprint()

i = 1
x = 1
total_a = 0
total_b = 0
total_c = 0
total_d = 0

for n in arraylist:
for z in range(len(n)-1)
value_a = int(a_value[i])
total_a+=value_a
i+= 1
x+=1









share|improve this question
















i'm struggling to work out how to use a loop to run through a list which contains 4 array values being "a_value", "b_value", "c_value", "d_value". Each of my arrays in this list have 23 elements. the outcome im looking for is for a loop to run though the list adding all 23 elements of each array. E.G. total of "a_value" then total of "b_value" and so on. this is what i have so far.
Thank you guys.



class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None

class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value) # each '""_value' has 23 integer values
in them
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

arraylist.listprint()

i = 1
x = 1
total_a = 0
total_b = 0
total_c = 0
total_d = 0

for n in arraylist:
for z in range(len(n)-1)
value_a = int(a_value[i])
total_a+=value_a
i+= 1
x+=1






python arrays linked-list nested-loops






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 14 '18 at 16:54









NoorJafri

583312




583312










asked Nov 14 '18 at 14:08









Chris DayChris Day

63




63












  • What is your input and expected output?

    – Alex
    Nov 14 '18 at 14:15











  • Did it help you?

    – NoorJafri
    Nov 14 '18 at 14:36

















  • What is your input and expected output?

    – Alex
    Nov 14 '18 at 14:15











  • Did it help you?

    – NoorJafri
    Nov 14 '18 at 14:36
















What is your input and expected output?

– Alex
Nov 14 '18 at 14:15





What is your input and expected output?

– Alex
Nov 14 '18 at 14:15













Did it help you?

– NoorJafri
Nov 14 '18 at 14:36





Did it help you?

– NoorJafri
Nov 14 '18 at 14:36












1 Answer
1






active

oldest

votes


















0














you can try something like this:



 a_value = [1,2,3]
b_value = [4,5,6]
c_value = [7,8,9]
d_value = [10,11,12]

class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None


class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value)
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

node = arraylist.headval # fetch head node
nextval = node.nextval # fetch next to node
list =
list.append(node.dataval) # enter all head list

while nextval: # this loop will break as nextval is none
node = node.nextval # node now becomes next and repeat
if hasattr(node, 'nextval'):
nextval = node.nextval
else:
nextval = None
list.append(node.dataval)

sum = 0

for x in list: # sum in the end or break into 23 for all separate
for y in x:
sum+=y

print(list, sum)

#OUTPUT [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]] 78


Works fine :)






share|improve this answer

























  • (nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

    – Chris Day
    Nov 14 '18 at 15:29











  • Just add a check before fetching the value.

    – NoorJafri
    Nov 14 '18 at 15:32











  • what kind of check would you use?

    – Chris Day
    Nov 14 '18 at 15:54











  • great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

    – Chris Day
    Nov 14 '18 at 18:44











  • You have to play around that :D learn guy and accept answer if it helped you :p

    – NoorJafri
    Nov 14 '18 at 19:23










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














you can try something like this:



 a_value = [1,2,3]
b_value = [4,5,6]
c_value = [7,8,9]
d_value = [10,11,12]

class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None


class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value)
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

node = arraylist.headval # fetch head node
nextval = node.nextval # fetch next to node
list =
list.append(node.dataval) # enter all head list

while nextval: # this loop will break as nextval is none
node = node.nextval # node now becomes next and repeat
if hasattr(node, 'nextval'):
nextval = node.nextval
else:
nextval = None
list.append(node.dataval)

sum = 0

for x in list: # sum in the end or break into 23 for all separate
for y in x:
sum+=y

print(list, sum)

#OUTPUT [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]] 78


Works fine :)






share|improve this answer

























  • (nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

    – Chris Day
    Nov 14 '18 at 15:29











  • Just add a check before fetching the value.

    – NoorJafri
    Nov 14 '18 at 15:32











  • what kind of check would you use?

    – Chris Day
    Nov 14 '18 at 15:54











  • great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

    – Chris Day
    Nov 14 '18 at 18:44











  • You have to play around that :D learn guy and accept answer if it helped you :p

    – NoorJafri
    Nov 14 '18 at 19:23















0














you can try something like this:



 a_value = [1,2,3]
b_value = [4,5,6]
c_value = [7,8,9]
d_value = [10,11,12]

class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None


class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value)
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

node = arraylist.headval # fetch head node
nextval = node.nextval # fetch next to node
list =
list.append(node.dataval) # enter all head list

while nextval: # this loop will break as nextval is none
node = node.nextval # node now becomes next and repeat
if hasattr(node, 'nextval'):
nextval = node.nextval
else:
nextval = None
list.append(node.dataval)

sum = 0

for x in list: # sum in the end or break into 23 for all separate
for y in x:
sum+=y

print(list, sum)

#OUTPUT [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]] 78


Works fine :)






share|improve this answer

























  • (nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

    – Chris Day
    Nov 14 '18 at 15:29











  • Just add a check before fetching the value.

    – NoorJafri
    Nov 14 '18 at 15:32











  • what kind of check would you use?

    – Chris Day
    Nov 14 '18 at 15:54











  • great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

    – Chris Day
    Nov 14 '18 at 18:44











  • You have to play around that :D learn guy and accept answer if it helped you :p

    – NoorJafri
    Nov 14 '18 at 19:23













0












0








0







you can try something like this:



 a_value = [1,2,3]
b_value = [4,5,6]
c_value = [7,8,9]
d_value = [10,11,12]

class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None


class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value)
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

node = arraylist.headval # fetch head node
nextval = node.nextval # fetch next to node
list =
list.append(node.dataval) # enter all head list

while nextval: # this loop will break as nextval is none
node = node.nextval # node now becomes next and repeat
if hasattr(node, 'nextval'):
nextval = node.nextval
else:
nextval = None
list.append(node.dataval)

sum = 0

for x in list: # sum in the end or break into 23 for all separate
for y in x:
sum+=y

print(list, sum)

#OUTPUT [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]] 78


Works fine :)






share|improve this answer















you can try something like this:



 a_value = [1,2,3]
b_value = [4,5,6]
c_value = [7,8,9]
d_value = [10,11,12]

class node():
def __init__(self, dataval = None):
self.dataval = dataval
self.nextval = None


class linked_list():
def __init__(self):
self.headval = None

def listprint(self):
printval = self.headval
while printval != None:
print (printval.dataval)
printval = printval.nextval


arraylist = linked_list()
arraylist.headval = node(a_value)
e2 = node(b_value)
e3 = node(c_value)
e4 = node(d_value)

arraylist.headval.nextval = e2
e2.nextval = e3
e3.nextval = e4

node = arraylist.headval # fetch head node
nextval = node.nextval # fetch next to node
list =
list.append(node.dataval) # enter all head list

while nextval: # this loop will break as nextval is none
node = node.nextval # node now becomes next and repeat
if hasattr(node, 'nextval'):
nextval = node.nextval
else:
nextval = None
list.append(node.dataval)

sum = 0

for x in list: # sum in the end or break into 23 for all separate
for y in x:
sum+=y

print(list, sum)

#OUTPUT [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]] 78


Works fine :)







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 14 '18 at 16:44

























answered Nov 14 '18 at 14:18









NoorJafriNoorJafri

583312




583312












  • (nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

    – Chris Day
    Nov 14 '18 at 15:29











  • Just add a check before fetching the value.

    – NoorJafri
    Nov 14 '18 at 15:32











  • what kind of check would you use?

    – Chris Day
    Nov 14 '18 at 15:54











  • great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

    – Chris Day
    Nov 14 '18 at 18:44











  • You have to play around that :D learn guy and accept answer if it helped you :p

    – NoorJafri
    Nov 14 '18 at 19:23

















  • (nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

    – Chris Day
    Nov 14 '18 at 15:29











  • Just add a check before fetching the value.

    – NoorJafri
    Nov 14 '18 at 15:32











  • what kind of check would you use?

    – Chris Day
    Nov 14 '18 at 15:54











  • great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

    – Chris Day
    Nov 14 '18 at 18:44











  • You have to play around that :D learn guy and accept answer if it helped you :p

    – NoorJafri
    Nov 14 '18 at 19:23
















(nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

– Chris Day
Nov 14 '18 at 15:29





(nextval = node.nextval) is getting error " 'Nonetype' object has no attribute 'nextval' "

– Chris Day
Nov 14 '18 at 15:29













Just add a check before fetching the value.

– NoorJafri
Nov 14 '18 at 15:32





Just add a check before fetching the value.

– NoorJafri
Nov 14 '18 at 15:32













what kind of check would you use?

– Chris Day
Nov 14 '18 at 15:54





what kind of check would you use?

– Chris Day
Nov 14 '18 at 15:54













great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

– Chris Day
Nov 14 '18 at 18:44





great! thank you buddy! just one last thing as im curious, is there a way to get it so it would print "[[6],[15],[24],[33]]" instead Thanks :)

– Chris Day
Nov 14 '18 at 18:44













You have to play around that :D learn guy and accept answer if it helped you :p

– NoorJafri
Nov 14 '18 at 19:23





You have to play around that :D learn guy and accept answer if it helped you :p

– NoorJafri
Nov 14 '18 at 19:23



















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