CODEWITHSUNDEEP
python
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clink

Reputation: 213

Joining two strings within a list of list

I have a list of lists containing:

animal = [[1, 'Crocodile','Lion'],[2, 'Eagle','Sparrow'],[3, 'Hippo','Platypus','Deer']]

and I want to join the string elements in each list inside the animal table so that it becomes a single string:

 animal = [[1, 'Crocodile, Lion'],[2, 'Eagle, Sparrow'],[3,'Hippo, Platypus, Deer']]

I tried using a for loop to join them:

for i in range(len(animal)):
     ''.join(animal[1:]) #string at index 1 and so on
print(animal)

I'm getting a type error saying "TypeError: sequence item 0: expected str instance, list found".

Upvotes: 2

Views: 268

Answers (7)

user9507356
user9507356

Reputation: 51

l1=[[1, 'Crocodile','Lion'],[2, 'Eagle','Sparrow'],[3, 'Hippo','Platypus','Deer']]

l2 = list(map(lambda x: [x[0],"{}, {}".format(x[1],x[2])],l1))
print(l2)

Upvotes: 0

aclown
aclown

Reputation: 101

Just a small change needed, you just forgot the index i in your loop:

animal = [[1, 'Crocodile','Lion'],[2, 'Eagle','Sparrow'],[3, 'Hippo','Platypus','Deer']]

merged_animal = []

for element in animal:
    merged_animal.append([element[0], ", ".join(element[1:])])

print(merged_animal)

But if you know list-comprehensions, better use them as shown in many answers.

Upvotes: 1

penguin2048
penguin2048

Reputation: 1343

Firstly, the type error is coming because you are applying join() on the list animal not on the sublists of list animal.

You should also keep in mind that join does not edit the original list, it just returns the new string.

Hence if you keep the above two things in mind your new code will look something like this

for i in range(len(animal)):
 animal[i] = [animal[i][0], ', '.join(animal[i][1:])] #string at index 1 and so on
print(animal)

the above code replaces each sublist with another sublist containing the sublist number and a string formed by joining the remaining part of original sublist with a ', ' (note your mistake here, you were joining with an empty character but your requirement is a comma and a space).

Upvotes: 0

Neil
Neil

Reputation: 796

animal is a list of lists so you need an extra index. You see this by adding 

print(animal[i])
print(animal[i][0])
print(animal[i][1])

etc in the loop

animal = [[1, 'Crocodile','Lion'],[2, 'Eagle','Sparrow'],[3, 'Hippo','Platypus','Deer']]

for i in range(len(animal)):
    print(animal[i][1:])
    animal[i][1] = ' '.join(animal[i][1:]) #string at index 1 and so on
    del animal[i][2:]


print(animal)

Upvotes: 0

timgeb
timgeb

Reputation: 78800

There are two problems with your code:

animal[1:] is the following list

>>> animal[1:]
[[2, 'Eagle', 'Sparrow'], [3, 'Hippo', 'Platypus', 'Deer']]

what do you expect to happen if you join it in every iteration of your loop? The second problem is that you do not assign the return value of join to anything, so even if the operation would not throw an error, you would lose the result.

Here's a solution with extended iterable unpacking:

>>> animal = [[1, 'Crocodile','Lion'],[2, 'Eagle','Sparrow'],[3,'Hippo','Platypus','Deer']]
>>> [[head, ', '.join(tail)] for head, *tail in animal]
[[1, 'Crocodile, Lion'], [2, 'Eagle, Sparrow'], [3, 'Hippo, Platypus, Deer']]

Here's one without:

>>> [[sub[0], ', '.join(sub[1:])] for sub in animal]
[[1, 'Crocodile, Lion'], [2, 'Eagle, Sparrow'], [3, 'Hippo, Platypus, Deer']]

Upvotes: 0

Maurice Meyer
Maurice Meyer

Reputation: 18136

>>> [[a[0], ','.join(a[1:])] for a in animal]
>>> [[1, 'Crocodile,Lion'], [2, 'Eagle,Sparrow'], [3, 'Hippo,Platypus,Deer']]

Upvotes: 1

Eric Duminil
Eric Duminil

Reputation: 54313

animal could be called table, and should indicate that it's not just a single animal. It also shouldn't be called animals, because it's not just a list of animals.

Thanks to unpacking, you can split your sub-lists directly into an integer and a list of animals:

>>> table = [[1, 'Crocodile','Lion'],[2, 'Eagle','Sparrow'],[3, 'Hippo','Platypus','Deer']]
>>> [[i, ', '.join(animals)] for (i, *animals) in table]
[[1, 'Crocodile, Lion'], [2, 'Eagle, Sparrow'], [3, 'Hippo, Platypus, Deer']]

Upvotes: 1

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