how to remove items from list which are present side by side in another list?

Question

grt=["thousand","million","billion","trillion","quadrillion","quintillion","sextillion","septillion","octillion","nonillion","decillion","undecillion","duodecillion","tredecillion","quattuor-decillion","quindecillion","sexdecillion","septen-decillion","octodecillion","novemdecillion","vigintillion"]


e=["sixty","million","billion","trillion","quadrillion","quintillion","sextillion","septillion","octillion"]

def sep():
    global e
    global grt

    for i in range(1,len(e)):
        if (e[i] in grt and e[i-1] in grt)==True:
            e.remove(e[i])

while all(((i in grt) and (e[e.index(i)-1] in grt)) for i in e[1:len(e)])!=False:
    sep()
print(e)

i am expecting that the last statement print(e) would print ["sixty"] but its not printing that.

...or `print([e[i] for i in range(len(e)) if e[i]!=grt[i]])` — Trapli, Oct 13 '19 at 12:31

score 0 · Answer 1 · answered Oct 13 '19 at 12:48

Filter function is perfect for such application.


grt=["thousand","million","billion","trillion","quadrillion","quintillion","sextillion","septillion","octillion","nonillion","decillion","undecillion","duodecillion","tredecillion","quattuor-decillion","quindecillion","sexdecillion","septen-decillion","octodecillion","novemdecillion","vigintillion"]


e=["sixty","million","billion","trillion","quadrillion","quintillion","sextillion","septillion","octillion"]

results = list(filter(lambda x:  x not in grt, e))
print(results)

how to remove items from list which are present side by side in another list?

1 Answers1