Print a result without square brackets

Question

When I create a list of arrays in Python and print out the result of the first index, why does it print the square brackets too?

    people = [
    ["Fredrick", 23],
    ["Lorenzo", 13],
    ["Francesco", 13],
    ["Giovanna", 9]
]

print(f"{people[0]}")

Output: ['Fredrick', 23]

`people[0]` is the list `['Fredrick', 23]`, and lists get printed with square brackets around, just like you enter them, so what surprises you here? Did you expect anything different? — Thierry Lathuille, Dec 23 '18 at 12:59
Welcome to Stack Overflow! Please read the [tour], and browse the [Help] to get an insight of how to ask a good question. I have edited your post title so it tells us about your actual question, instead of your own generic title. — Jongware, Dec 23 '18 at 15:26

score 1 · Answer 1 · answered Dec 23 '18 at 13:31

You created people which is a list of lists. Then you refer to the f-string f"{people[0]}". When you reference it, the expression people[0] is evaluated and is found to be the list ["Fredrick", 23]. That is then converted to a string--remember, this is an f-string. The double-quotes around Fredrick are converted to single-quotes, according to the Python standard, where either single- or double-quotes may be used to denote a string but they are alternated for strings-within-strings. Python wants to put double-quotes around the entire string whenever its representation is needed, so to alternate the type of quotes, single-quotes are used. So the final f-string is

"['Fredrick', 23]"

Remember, what is inside the string is a representation of the list inside people. Python adds the brackets so everyone knows that is indeed a list. Python could have done it differently, but this makes the list nature explicit, and the second rule in the Zen of Python is

Explicit is better than implicit.

So the brackets are shown. If you do not want the brackets, there are several ways to remove them. The simplest way to print that f-string without the brackets is probably

print(f"{people[0]}"[1:-1])

which prints

'Fredrick', 23

That code prints the string but removes the first and last characters, which are the brackets. If you do not understand Python's slice notation you should research it--this is one of Python's strengths in string handling.

Thank you, it was really helpful. – wallen Dec 23 '18 at 13:51 — wallen, Dec 23 '18 at 13:51

Daedalus · Accepted Answer · 2018-12-23T15:21:40.790

1

You defined an array with arrays as elements. That's why you get the array ['Fredrick', 23] when you access the first index of people array. This is equivalent with accessing the first item of the array

>> array = [10, 20, 30]
>> print array[0]
>> 10

In your case the first element is the array:

>> ['Fredrick', 23]

An alternative approach could be:

# map is another (more appropriate) 
# data structure to store this data. 
people = {
    'Fredrick': 23,
    'Lorenzo': 13,
    'Francesco': 13,
    'Giovanna': 9
}

for name, age in people.items():
    # iterate through all map items
    # and print out each person's name and age.
    # Note: you need to convert the age from int to string
    # in order to concatenate it.
    print name + ", " + str(people[name])

edited Dec 23 '18 at 15:21

answered Dec 23 '18 at 14:54

Daedalus

295
2
17

people_map.items() is not defined ? – wallen Dec 23 '18 at 15:15
In my local environment I used people_map instead of people and I forgot to update this line of code. Sorry! Now its fixed – Daedalus Dec 23 '18 at 15:23

score 0 · Answer 3 · answered Dec 23 '18 at 12:59

people = [
    ["Fredrick", 23],
    ["Lorenzo", 13],
    ["Francesco", 13],
    ["Giovanna", 9]
]

Creates a list of lists, people[0] gives the first value - which is the list ["Fredrick", 23]. When printing a list python (by default) print the brackets as well.

Print a result without square brackets

3 Answers3