Is there a way to write a for each in python3 that modifies the array?

Question

Let's say for example that I have an array that represents a name and I want to use just the first letter of any part of a name that is not the first or the last part:

# Have
name = ["John", "Banana", "Doe"]

# Want
["John", "B", "Doe"]

Can I iterate through a subset of an array and change the variables inside it through the variable that references it like so?

for part in name[1:-1]:
    part = part[0]

The only solution I fould for this was to use list comprehention, but it was too hard to read:

name = [part if i in [0, len(name)-1] else part[0] for i, part in enumerate(name)]

comprehension can indeed be confounding. maybe bury it in an appropriately named function or use comments. you could also use numpy arrays of dtype object instead of lists for an alternate syntax, but additional dependencies is worse than keeping the confoundation. — Abel
– Abel, Commented Feb 15, 2022 at 13:15

timgeb · Accepted Answer · 2022-02-15 13:29:15Z

4

v = v[0] reassigns the name v, that was previously bound to one of the list elements. This does not mutate the list, though.

You could fix your loop by iterating over an index range and explicitly reassign list elements.

for index in range(1, len(name) - 1):
    name[index] = name[index][0]

An easier to read list-comprehension solution would be:

name[1:-1] = [v[0] for v in name[1:-1]]

An equivalent solution without a list comprehension would be:

new_values = []

for v in name[1:-1]:
    new_value = v[0]
    new_values.append(new_value)

name[1:-1] = new_values

Bonus: repeating [1:-1] twice is not as clean as it could be. You could define a slice object to refer to the desired part of the list, e.g.

where = slice(1, -1)
name[where] = [v[0] for v in name[where]]

edited Feb 15, 2022 at 13:29

answered Feb 15, 2022 at 13:15

timgeb

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3 Comments

Scott Hunter Over a year ago

What is the point of building up a list to assign to a slice when you can just do the individual assignments directly (as in your first solution)?

timgeb Over a year ago

@ScottHunter educational reasons. OP is new to python and finds list comprehension hard to read.

Scott Hunter Over a year ago

I thought OP meant that comprehension in particular; yours was about as straight-forward as one could get.

Scott Hunter · Accepted Answer · 2022-02-15 13:15:36Z

1

start = 1
for i,v in enumerate(name[start:-1]):
    name[i+start] = v[0]

Or something slightly cleaner, using what I just learned about enumerate:

start = 1
for i,v in enumerate(name[start:-1],start):
    name[i] = v[0]

answered Feb 15, 2022 at 13:15

Scott Hunter

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Comments

mozway · Accepted Answer · 2022-02-15 13:14:27Z

0

What about using enumerate with a set?

name = ["John", "Banana", "Doe"]

pos = {0, len(name)-1}

[w if i in pos else w[0] for i,w in enumerate(name)]

or with a classical loop (but without explicit check, the list should have more than 1 element):

out = name[:1]

for i in range(1, len(name)-1):
    out.append(name[i][0])

out.append(name[-1])

answered Feb 15, 2022 at 13:14

mozway

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Comments

Mohamed ElKalioby · Accepted Answer · 2022-02-15 13:13:07Z

-1

You can do this

for i,v in enumerate(name[1:-1]):
   name[i] = v[0]

What you can't do while iterating over a list is changing the list size, but changing the content is allowed.

answered Feb 15, 2022 at 13:13

Mohamed ElKalioby

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11 Comments

Scott Hunter Over a year ago

Not if you want OP's desired result.

Mohamed ElKalioby Over a year ago

What do you mean by that?

Scott Hunter Over a year ago

Did you actually try this & check the results?

jackal Over a year ago

That doesn't seem to work

João Fukuda Over a year ago

Another problem with this approach is that the i starts at 0, but the iteration starts at 1. You would need to use name[i+1] to make it work.

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