Simple if elif mismatched result

Question

This if-elif is simple and straightforward. index is a 1-D array with values of 0-5 only. As you can see from the image, the only correct if-elif is only for index[i]==0 and index[i]==1. For index[i]==5, it was supposed to print f but the result was printed as d. What went wrong?

Current output:

for i in index:
    print(i)
    if index[i]==0:
      print(" :a")
    elif index[i]==1:
        print(" :b")
    elif index[i]==2:
        print(" :c")
    elif index[i]==3:
        print(" :d")
    elif index[i]==4:
        print(" :e")
    elif index[i]==5:
        print(" :f")

i mean, what is value, you have assigned to index in for loop? — BugHunter, May 05 '18 at 13:09
`i` loops over _values_, not indices of `index`! You just need `i` instead of `index[i]`. — Andras Deak -- Слава Україні, May 05 '18 at 13:11
Probably not the best dupe target but this is basic python behaviour. — Andras Deak -- Слава Україні, May 05 '18 at 13:13
But if you have a scalar index from 0 to 5 you can drop the ifs and use `'abcdef'[i]`. — Andras Deak -- Слава Україні, May 05 '18 at 13:16
How do I do that? Is it as below? for i in index: print(i) 'abcdef'[i] — Projia, May 05 '18 at 14:01

Pitto · Answer 1 · 2018-05-08T12:12:33.460

0

I've tried your code and the issue is related to the fact that in each loop you have some confusion between the index and the value.

Here, using enumerate, I can access both index and value at every loop:

index_list = [10,4,2,3,4,15,7]

for index, value in enumerate(index_list):
    print "\nCurrent index: " + str(index)
    print "Current value: " + str(value)
    print "Current result:"
    if value==0:
        print(" :a")
    elif value==1:
        print(" :b")
    elif value==2:
        print(" :c")
    elif value==3:
        print(" :d")
    elif value==4:
        print(" :e")
    elif value==5:
        print(" :f")
    else:
        print("This value is not in the 0 - 5 range, skipping it...")

edited May 08 '18 at 12:12

answered May 05 '18 at 13:12

Pitto

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You're just making the same mistake as OP. Try with `index = [10,11,12]` etc. – Andras Deak -- Слава Україні May 05 '18 at 13:14
Correct indeed! Should be working now – Pitto May 05 '18 at 13:20
Nope, sorry. As I said, try with test values starting from 10. – Andras Deak -- Слава Україні May 05 '18 at 13:22
@AndrasDeak The OP is writing: "with values of 0-5 only" – Pitto May 05 '18 at 13:24
The code still only doesn't fail by accident :) Like driving on the wrong side of the highway when there's no other car on the road. – Andras Deak -- Слава Україні May 05 '18 at 13:31
I don't get it... How can you make the code fail if the provided array is only as in the requirements? – Pitto May 05 '18 at 13:38
1

@AndrasDeak I think I've understood :D Thanks :) – Pitto May 05 '18 at 13:44
Can you spot a difference between your code and OP's code? That's the error! – Austin May 05 '18 at 13:45
@theausome It should be ok now :) – Pitto May 05 '18 at 13:47
1

Yup. Now only a style remark: use `val` everywhere if you have it :) – Andras Deak -- Слава Україні May 05 '18 at 14:00
1

Fully agreed, @AndrasDeak. I've now improved the style as you pointed out and added a few debug prints. My pig now has lipstick too :D Thanks for the suggestions and the patience. – Pitto May 08 '18 at 07:09

Austin · Accepted Answer · 2018-05-05T14:04:16.917

0

You can just shorten your code by avoiding those if elifs using a dictionary to map to required values:

index = [1, 5, 5, 5, 5, 4, 4, 4, 0]
map_dict = {0: "a", 1: "b", 2: "c", 3: "d", 4: "e", 5: "f"}

for i in index:
    print(map_dict.get(i))

# b
# f                                                        
# f                                                      
# f                                                        
# f                                                         
# e                                                  
# e                                                      
# e                                                       
# a

EDIT:

To get count of each of the item in the output:

from collections import Counter

index = [1, 5, 5, 5, 5, 4, 4, 4, 0]
map_dict = {0: "a", 1: "b", 2: "c", 3: "d", 4: "e", 5: "f"}

lst = []
for i in index:
    value = map_dict.get(i)
    print(value)
    lst.append(value)

print(Counter(lst))

edited May 05 '18 at 14:04

answered May 05 '18 at 13:23

Austin

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is it possible to count the totals of each 'a', 'b','c','d', 'e' and 'f' of the output? because in if elif, i can just use countera+=1 etc – Projia May 05 '18 at 13:57
@Projia `collections.Counter(index)` to get the counts of each index, then use the above dict to get the count of characters. – Andras Deak -- Слава Україні May 05 '18 at 14:03
See the edit now. – Austin May 05 '18 at 14:04
2

Thanks @AndrasDeak and theausome. It seems that dictionaries are very convenient and straightforward. Will learn more on it. – Projia May 05 '18 at 14:25

Simple if elif mismatched result

2 Answers2