zip a list with space delimited into tuple

Question

I have a dataframe df with a column called columnList as str.

"1 2,7 8,10 7"

Then I converted them to a list as shown :

[1 2,7 8,10 7]

I want to convert the value inside list to tuple :

[(1,2),(7,8),(10,7)]

Current code :

temp = df['columnList'].str.split(',') result = list(zip(temp[::2], temp[1::2])) print(result)

I'm getting empty list.

df looks like this :

column1 columnList YY 1 2,7 8,10 7

Name: df, dtype: object

PacketLoss · Accepted Answer · 2020-06-02 02:29:26Z

You don't need to use zip here, just iterate over the list, split each element and store it as a tuple.

l = [ '1 2', '7 8', '10 7'] [tuple(int(i) for i in numbers.split()) for numbers in l] #[(1, 2), (7, 8), (10, 7)]

sushanth · Accepted Answer · 2020-06-02 03:25:00Z

1

try this,

df.columnsList.apply(lambda x : [tuple(map(int, x.split())) for x in "1 2, 7 8, 10 7".split(",")])

output,

0 [(1, 2), (7, 8), (10, 7)] Name: columnsList, dtype: object

answered Jun 2, 2020 at 3:02

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2 Comments

It works but the numbers are not converted to float. It is in string

This works too : apply(lambda x : [tuple(map(float,x.split())) for x in x.split(",")]) Thank you!

adamgy · Accepted Answer · 2020-06-02 02:31:12Z

You could map the characters to ints after splitting them, then convert the map object to a tuple:

temp = df['columnList'].str.split(',') result = [tuple(map(int, num.split())) for num in temp] print(result) # [(1, 2), (7, 8), (10, 7)]