Python 根据列表元素第一个数字的相等性将列表分成子列表

Question

a = [58,23,12,24,11,63,54]

I want to make the above list as follows:

[[58,54],[23,24],[12,11]]

ie, I want to make the list of sublists whose elements start with same digit and having count > 1.

Answer 1

You can use dictionary

from collections import defaultdict
a = [58,23,12,24,11,63,54]

d = defaultdict(list)
for num in a:
    d[str(num)[0]].append(num)
    
res = [v for v in d.values() if len(v) > 1]
print(res)

Output:

[[58, 54], [23, 24], [12, 11]]

Answer 2

This code groups integer of any digit on the basis of the first digit. I hope this is what you are looking for. Also this may not be the most efficient way of doing this.

a=[58,23,12,24,11,63,54]
final=[[],[],[],[],[],[],[],[],[],[]]
for i in a:
    count=0
    a=i
    while (i>0):
        count+=1
        i=i//10
    final[int(a/(10**(count-1)))].append(a)
a=[]
for i in final:
    if len(i)>1:
        a.append(i)
print(a)

Answer 3

you can use itertools.groupby to group a list by any key function you want:

note the use of Python 3.8's "walrus operator"

from itertools import groupby

a=[58,23,12,24,11,63,54]
get_first_digit = lambda n: str(n)[0]
sorted_by_digit = sorted(a, key=get_first_digit, reverse=True)  # sorting is a must for groupby
result = [y for _,g in groupby(sorted_by_digit, key=get_first_digit) if len(y:=list(g))>1]

print(result)

Output:

[[58, 54], [23, 24], [12, 11]]

Answer 4

I have an answer to this, although it is very big.

a = [58,23,12,24,11,63,54]
s = [[a[0]]]

def firstdigit(n):
    while n>=10:
        n = n//10
    return n

def arrange(s, i):
    for j in range(len(s)):
        if firstdigit(s[j][0]) == firstdigit(i):
            s[j].append(i)
            break
    else:
        m=[]
        m.append(i)
        s.append(m)
    return s

def check(s):
    for i in s:
        if len(i) <= 1:
            s.remove(i)
    return s

for i in a[1:]:
    s = arrange(s, i)

#checks for count>1
s = check(s)
print(s)

This code contains only basics with no modules imported.