Efficiently Zip multiple list in python/pandas

Question

I have below Lists

Number = [[1],[2],[3],[4],[6]]
L1 = ['A','B','C','D','E']
L2 = [100, 55, 315, 68, 23]
L3 = ['18%','105','56%','12%','4%']

I wanted to Zip all the lists and create a DataFrame. I used the below code and successfully able to do it.

for n, l1, l2, l3 in zip(Number,L1,L2, L3):
    n.insert(1,l1)
    n.insert(2,l2)
    n.insert(3,l3)

df = pd.DataFrame(Number, columns=['Number','Name', 'Value', 'Score'])
print(df)

+---+--------+------+-------+-------+
|   | Number | Name | Value | Score |
+---+--------+------+-------+-------+
| 0 |    1   |   A  |  100  |  18%  |
+---+--------+------+-------+-------+
| 1 |    2   |   B  |   55  |  105  |
+---+--------+------+-------+-------+
| 2 |    3   |   C  |  315  |  56%  |
+---+--------+------+-------+-------+
| 3 |    4   |   D  |   68  |  12%  |
+---+--------+------+-------+-------+
| 4 |    6   |   E  |   23  |   4%  |
+---+--------+------+-------+-------+

Since there is only 4 lists in this example. Easily we can type manually for n, l1, l2, l3 in zip(Number,L1,L2, L3): and type individual insert functions. Now my question is, what if there is many lists (say 15)? is there a pythonic way of doing this?

Does this answer your question? [Python: create a pandas data frame from a list](https://stackoverflow.com/questions/43175382/python-create-a-pandas-data-frame-from-a-list) — David Erickson, Jul 31 '20 at 17:06
No. I'm asking about efficient way of Zipping multiple files. the above is different. — Sunni, Jul 31 '20 at 17:08
where is the mention of anything about zipping multiple `files` in your question? — David Erickson, Jul 31 '20 at 17:09
you can try `pd.DataFrame(zip(np.concatenate(Number),L1,L2,L3))` — anky, Jul 31 '20 at 17:11
Or `pd.DataFrame(zip(chain.from_iterable(Number),L1,L2,L3))`. — Henry Yik, Jul 31 '20 at 17:13
@Sunni I don't see a mention of anything regarding `multiple files` — David Erickson, Jul 31 '20 at 17:13
@DavidErickson it is not about multiple `files`. it's about `zipping` multiple `lists`.. — Sunni, Jul 31 '20 at 17:14
Okay @Sunni there was confusion, because you misspoke in this comment... "No. I'm asking about efficient way of Zipping `multiple files`. the above is different". Anyway, I would go with anky or Henry Yik's answer. — David Erickson, Jul 31 '20 at 17:20
@DavidErickson. No worries. cool. Yes agreed I prefer anky's answer too — Sunni, Jul 31 '20 at 17:33
To close the loop, the first comment link I shared may not be a match, but I think this SO question / answers is a match: https://stackoverflow.com/questions/30522724/take-multiple-lists-into-dataframe — David Erickson, Jul 31 '20 at 17:41

jsmart · Answer 1 · 2020-07-31T17:14:09.153

0

Does this meet the requirement? Make a dictionary with the name of each list, and the list itself. Then pass the dictionary to the DataFrame constructor (in the print statement below).

data = {
    'Number': Number,
    'L1': L1,
    'L2': L2,
    'L3': L3
}
df = pd.DataFrame(data).rename(columns={'L1': 'Name', 
                                        'L2': 'Value', 
                                        'L3': 'Score'})
print(df)

   Number Name  Value Score
0       1    A    100   18%
1       2    B     55   105
2       3    C    315   56%
3       4    D     68   12%
4       6    E     23    4%

edited Jul 31 '20 at 17:14

answered Jul 31 '20 at 17:08

jsmart

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this answer doesn't work. because, `Number` is a `list of lists`. first we have to flatten this list by `[item for sublist in Number for item in sublist]`. after doing this, we can use your code – Sunni Feb 20 '21 at 11:45

score 0 · Answer 2 · edited Jul 31 '20 at 17:14

0

I would definitely find a more resumed format to write this query! Out of the top of my head, I thought about this:

test = [Number, L1, L2, L3]
for t in zip(*test):
    print(t)

# on the first loop iteration:
# t = ([1], 'A', 100, '18%')
#t[3] = ['18%']

As you'll find, your results will now be in a tuple (e.g. ([1], 'A', 100, '18%')). So, if you need to access each value, e.g. the value from Number, you can do t[0].

edited Jul 31 '20 at 17:14

Dharman

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answered Jul 31 '20 at 17:09

rmssoares

175
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`Number` is a list of list – Sunni Jul 31 '20 at 17:35

score 0 · Answer 3 · answered Jul 31 '20 at 17:20

0

I am not sure why you have Number be a list of lists. But, this may be what you're looking for:

Number = [1,2,3,4,6]
L1 = ['A','B','C','D','E']
L2 = [100, 55, 315, 68, 23]
L3 = ['18%','105','56%','12%','4%']

pd.DataFrame(list(zip(Number, L1, L2, L3)), columns=['Number', 'Name', 'Value', 'Score'])

output:

   Number Name  Value Score
0       1    A    100   18%
1       2    B     55   105
2       3    C    315   56%
3       4    D     68   12%
4       6    E     23    4%

answered Jul 31 '20 at 17:20

aggis

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`Number` is a list of list – Sunni Jul 31 '20 at 17:35
It seems like you want to flatten the list then – aggis Jul 31 '20 at 17:37

Efficiently Zip multiple list in python/pandas

3 Answers3