如何在多个字符串的每个位置找到最频繁出现的字符

Question

我有一个不同长度的单词列表。
我想在所有单词的每个 position 中找到出现频率最高的字符。 这样做并防止错误index out of range的有效方法是什么？
例如：

alist = ['fowey', 'tynemouth', 'unfortunates', 'patroness', 'puttying', 'presumptuousness', 'lustrous', 'gloxinia']

所有单词的每个 position 中出现频率最高的字符（ 0 to len(max(alist, key=len)) ）等于： poternusakesness

for p in range (len(max(words, key=len))):

那么，如果两个字符具有相同的频率，第一个字符（基于字母表）被选为最常见的字符呢？

Answer 1

尝试：

from collections import Counter

alist = [
    "fowey",
    "tynemouth",
    "unfortunates",
    "patroness",
    "puttying",
    "presumptuousness",
    "lustrous",
    "gloxinia",
]

for i in (
    (w[idx] if idx < len(w) else None for w in alist)
    for idx in range(len(max(alist, key=len)))
):
    print(
        min(
            Counter(ch for ch in i if ch is not None).items(),
            key=lambda k: (-k[1], k[0]),
        )[0]
    )

印刷：

p
u
t
e
r
n
u
s
a
o
e
s
n
e
s
s

---

编辑：使用我们的collections.Counter ：

alist = [
    "fowey",
    "tynemouth",
    "unfortunates",
    "patroness",
    "puttying",
    "presumptuousness",
    "lustrous",
    "gloxinia",
]


for i in (
    (w[idx] if idx < len(w) else None for w in alist)
    for idx in range(len(max(alist, key=len)))
):
    cnt = {}
    for ch in i:
        if ch is not None:
            cnt[ch] = cnt.get(ch, 0) + 1

    print(
        min(
            cnt.items(),
            key=lambda k: (-k[1], k[0]),
        )[0]
    )