How to print all elements after a specific element in a list?-CodePudding

I want my output like if I choose the element 5 (not the index), only 4,7,6 will be printed.

b = [1,3,2,5,4,7,6]
Node = 5
for i in b:
    if Node == b[i-1]:
        continue;
    print(b[i-1])

CodePudding user response：

You could use the index method and then use the found index to slice the list:

b = [1,3,2,5,4,7,6]
Node = 5
index = b.index(Node)
for i in b[index 1:]:
    print(i)

CodePudding user response：

def main():
    b = [1, 3, 2, 5, 4, 7, 6]
    node = 5
    for i in range(-1, -len(b), -1):
        if b[i] == node:
            for j in b[i 1:]:
                print(j)
            return 0
    for i in b:
        print(i)


if __name__ == "__main__":
    main()

CodePudding user response：

b = [1, 3, 2, 5, 4, 7, 6]
Node = 5
print(b[b.index(Node) 1::])  # [4, 7, 6]

CodePudding user response：

b = [1,3,2,5,4,7,6]
Node=5
index=b.index(Node)# fetching index of node
for i in b[index 1::]: #iterating through list after the node
     print(i)

CodePudding user response：

To avoid slicing and creating new lists, you can create an iterator over the list and consume it until you reach the element, then print what is left:

l = [1, 3, 2, 5, 4, 7, 6]
node = 5

it = iter(l)
for num in it:
    if num == node:
        break

for num in it:
    print(num)

Or use itertools.dropwhile to avoid the first loop:

import itertools

l = [1, 3, 2, 5, 4, 7, 6]
node = 5

it = itertools.dropwhile(lambda x: x != node, l)
next(it)
for num in it:
    print(num)