Interleave Multiple Lists of the Same Length in Python

Interleave multiple lists of the same length in Python

Having posted the question, I've realised that I can simply do the following:

[val for pair in zip(l1, l2) for val in pair]

where l1 and l2 are the two lists.

---

If there are N lists to interleave, then

lists = [l1, l2, ...]
[val for tup in zip(*lists) for val in tup]

Interleave 4 lists of same length python

Provided the lists are the same length, zip() can be used to interleave four lists just like it was used for interleaving two in the question you linked:

>>> l1 = ["a", "b", "c", "d"]
>>> l2 = [1, 2, 3, 4]
>>> l3 = ["w", "x", "y", "z"]
>>> l4 = [5, 6, 7, 8]
>>> l5 = [x for y in zip(l1, l2, l3, l4) for x in y]
>>> l5
['a', 1, 'w', 5, 'b', 2, 'x', 6, 'c', 3, 'y', 7, 'd', 4, 'z', 8]

Interleave multiple lists of variable-length in Python3

Looks like you need itertools.zip_longest

from itertools import zip_longest
l1=[1,2,3]
l2=[10,20,30]
l3=[101,102,103,104]

print([y for x in zip_longest(l1,l2,l3, fillvalue=None) for y in x if y is not None])

Output:

[1, 10, 101, 2, 20, 102, 3, 30, 103, 104]

How to interleave two lists of different length?

def twolists(list1, list2):
    newlist = []
    a1 = len(list1)
    a2 = len(list2)

    for i in range(max(a1, a2)):
        if i < a1:
            newlist.append(list1[i])
        if i < a2:
            newlist.append(list2[i])

    return newlist

Interleaving Lists in Python

One option is to use a combination of chain.from_iterable() and zip():

# Python 3:
from itertools import chain
list(chain.from_iterable(zip(list_a, list_b)))

# Python 2:
from itertools import chain, izip
list(chain.from_iterable(izip(list_a, list_b)))

Edit: As pointed out by sr2222 in the comments, this does not work
well if the lists have different lengths. In that case, depending on
the desired semantics, you might want to use the (far more general) roundrobin()
function from the recipe
section of the
itertools documentation:

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).next for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

How to elegantly interleave two lists of uneven length in python?

if a is the longer list and b is the shorter

from itertools import groupby

len_ab = len(a) + len(b)
groups = groupby(((a[len(a)*i//len_ab], b[len(b)*i//len_ab]) for i in range(len_ab)),
                 key=lambda x:x[0])
[j[i] for k,g in groups for i,j in enumerate(g)]

eg

>>> a = range(8)
>>> b = list("abc")
>>> len_ab = len(a) + len(b)
>>> groups = groupby(((a[len(a)*i//len_ab], b[len(b)*i//len_ab]) for i in range(len_ab)), key=lambda x:x[0])
>>> [j[i] for k,g in groups for i,j in enumerate(g)]
[0, 'a', 1, 2, 'b', 3, 4, 5, 'c', 6, 7]

You can use this trick to make sure a is longer than b

b, a = sorted((a, b), key=len)

How to force three lists to have the same length in Python 3.7?

You could use these functions:

def padlist(lst, size, defvalue=None):
    if len(lst) < size:
        lst.extend([defvalue] * (size - len(lst)))

def padlists(lsts, defvalue=None):
    size = max(len(lst) for lst in lsts)
    for lst in lsts:
        padlist(lst, size, defvalue)

# Demo
lsts = [
    ["a", "b", "c"],
    ["a", "b", "c", "d", "e"],
    ["a", "b"],
]

padlists(lsts, "")  # provide the value to pad with

print(lsts)  # shorter lists are padded with blanks

How to interleave two lists by chunks

What about this:

A = [1.1, 1.1, 1.1, 1.1, 1.2, 1.2, 1.2, 1.2, 1.3, 1.3]
B = [2.1, 2.1, 2.1, 2.1, 2.2, 2.2, 2.2, 2.2, 2.3, 2.3]

sep = 4
num = len(A)
iterations = int(num / sep) + 1

merged = []
for i in range(iterations):
    start = sep * i
    end = sep * (i + 1)
    merged.extend(A[start:end])
    merged.extend(B[start:end])

print(merged)

>>> '[1.1, 1.1, 1.1, 1.1, 2.1, 2.1, 2.1, 2.1, 1.2, 1.2, 1.2, 1.2, 2.2, 2.2, 2.2, 2.2, 1.3, 1.3, 2.3, 2.3]'

Writing multiple lists of different lengths to same rows in Python with each list into a single cell

You can convert data to dictionary and append() to existing DataFrame

It will need to convert list [1, 2, 3] (and similar) to string "1,2,3" (etc.)

import pandas as pd

df = pd.DataFrame(columns=['Name', 'Date', 'Patcit', 'Npatcit', 'Class'])

# -------------------------------

Name = ['Jon']
Date = [1985]
Patcit = [1, 2, 3]
Npatcit = [4, 5, 6, 7, 8]
Class = [9, 10]

row = {
    'Name': Name[0],
    'Date': Date[0],
    'Patcit':  ','.join(str(x) for x in Patcit),
    'Npatcit': ','.join(str(x) for x in Npatcit),
    'Class':   ','.join(str(x) for x in Class),
}

df = df.append(row, ignore_index=True)

#print(df)

# -------------------------------

Name = ['Nikhil']
Date = [1988]
Patcit = [1, 2, 3]
Npatcit = [4, 5, 6, 7]
Class = [9, 10, 11, 12, 13]

row = {
    'Name': Name[0],
    'Date': Date[0],
    'Patcit':  ','.join(str(x) for x in Patcit),
    'Npatcit': ','.join(str(x) for x in Npatcit),
    'Class':   ','.join(str(x) for x in Class),
}

df = df.append(row, ignore_index=True)

print(df)

Result

     Name  Date Patcit    Npatcit          Class
0     Jon  1985  1,2,3  4,5,6,7,8           9,10
1  Nikhil  1988  1,2,3    4,5,6,7  9,10,11,12,13

And later you can write to csv using standard separator - comma - or other separator.

df.to_csv('output.csv', sep=';')

Or see other question which describes how to write fixed-width-file

---

Related Topics