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Python Python Sequences Sequence Operations Code Samples: Membership Testing, Count, and Index

Why does nums.index(10) show a ValueError: 10 is not in list?

I'm very confused with this.

nums = range(1, 10, 2)

nums.index(5) # 2 nums.index(10) # ValueError: 10 is not in list nums.index(1) # 0

Why does nums.index(5) show a 2 when it's not in the tuple? Why does nums.index(10) show a value error?

2 Answers

imo
imo
7,897 Points
# this does not really create a list
nums = range(1, 10, 2)
# if you print nums
# you will get
range(1, 10, 2)

# if you want to get a list from range(1, 10, 2), you will need to wrap this in a list(), like so
nums = list(range(1, 10, 2))
# ^ this will now return this when you try to print nums:
[1, 3, 5, 7, 9]
# ^ this makes sense, since range says to start at 1, skip every 2, until 10
# so all odd numbers from 1 - 10, which are 1, 3, 5, 7, 9

# Now, the reason why you where getting 2 when you typed:
nums.index(5)
# is because if you look at the list, [1, 3, 5, 7, 9]
# the number 5 is in the 2 index of that list
# so what .index(X) does, it look at the list for whatever number is passed in
# and it returns the index where that number is located

# The reason why you were getting ValueError when you typed:
nums.index(10)
# is because 10 is not in the list [1, 3, 5, 7, 9]
# if you did
nums.index(1) # you would get 0
nums.index(3) # you would get 1
nums.index(5) # you would get 2
nums.index(7) # you would get 3
nums.index(9) # you would get 4


# hope that helps
Joseph Yhu
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Joseph Yhu
PHP Development Techdegree Graduate 47,171 Points
  1. You are starting at 1 and incrementing by 2, so only the odd numbers would be returned. Also the stop value in the range function isn't included anyway.
  2. Remember the indices start at 0, so nums[0] = 1, nums[1] = 3, and nums[2] = 5.