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Good afternoon!
I have the list:
a=[4,3, 1, 2, 3]
and the list:
sub_ind_novo=[96, 7, 97, 23, 18]
And I needed to link the lists by getting the result:
no97. G1 #when sub_ind_novo
is 97, the corresponding in a is 1, so just print this 1 time
I need the sub_ind_new element printed as the number of times of the corresponding element in a.
(from here, when it takes values other than 1 in a I don’t know how to do.. that is, the 1st element in a is 4, and the respective sub_ind_new is 96, so it should appear 4 lines of this value in the way I write here below in the output example)
The rest should be in this form (example of result):
no23. G1
no23. G2
no7. G1
no7. G2
no7. G3
no18. G1
no18. G2
no18. G3
no96. G1
no96. G2
no96. G3
no96. G4
And I made the following code but I can’t make it for the times when it’s different from 1, that is, where the (2, 3, 4 or 5)..
for i in range(0,len(a)):
if a[i]==1:
print ("no%d. G1 \n" %sub_ind_novo[i])
elif:
.....
print ("no%d. G%d \n" %(sub_ind_novo[i],))
Thank you!
What language is this that Voce speaks?
– user45474
Voce could make a simpler analogy of your problem?
– user45474
I’m working on python 2.7
– Sofia Raimundo
You need to explain better. You want to know what the values of
a
which are equal to 1, and go to the respective index where you found them insub_ind_novo
? In this case he submitted he wants to withdraw the figuressub_ind_novo[11]
andsub_ind_novo[7]
?– Miguel
I’ve already edited, I’ve shortened the list size to make it easier. No, the "a" list is like a list of occurrences of the respective values of "sub_ind_new", so you should print the "sub_ind_new" element the number of times the respective element in "a", as in the result example. (96 has to appear 4 times, in the form: G1, G2, G3, G4; the 23 twice in the form: G1, G2, etc.. )
– Sofia Raimundo
very confusing, puts a larger part of the code and tries to explain better your problem
– Joannis
I’ve already worked it out, thank you!
– Sofia Raimundo