字母出现列表 Python

Question

我有一个编码任务，我需要创建一个程序来检查名称字典中包含字母 R、M、A、K 和 C 的次数。所以在这种情况下，我有这个字典：

dict = {"Renata", "Maciek", "Marek", "Karolina", "Marcel}

我需要创建一个循环来逐一检查字典中所有名称中的所有这些字母。完成后，它需要将结果放入一个列表中，并在程序结束时返回该列表。所以结果看起来像

R - 4 米 - 2 A - 3.（数字只是举例） K - 5 C - 6

这是我目前所拥有的：

names = {"Renata", "Marek", "Maciek", "Karolina", "Marcel"}

letter = ['R', 'M', 'A', 'K', 'C'] 

counter = 0 

for letter in names:

    print("R =", names.count("R"))
    print("M =", names.count("M"))
    print("A =", names.count("A"))
    print("K =", names.count("K"))
    print("C = ", names.count("C"))

如您所见，我对 Python 非常陌生，因此我非常感谢您的意见或帮助。请显示您对代码所做的更改。提前谢谢你

Answer 1

例如，这样：

names = {"Renata", "Marek", "Maciek", "Karolina", "Marcel"}
names_concatenated = "".join(names).upper()
letters = ['R', 'M', 'A', 'K', 'C']

for letter in letters:
    print(f"Letter '{letter} occurrences count in `names`: "
          f"{names_concatenated.count(letter)}")

我们有：

Letter 'R' occurrences count in `names`: 4
Letter 'M' occurrences count in `names`: 3
Letter 'A' occurrences count in `names`: 7
Letter 'K' occurrences count in `names`: 3
Letter 'C' occurrences count in `names`: 2

Answer 2

内置collections 库中的Counter 子类对于这类事情非常方便。我会这样实现它：

from collections import Counter

names = {"Renata", "Marek", "Maciek", "Karolina", "Marcel"}
all_letters = ''.join(names).upper()

counter = Counter(all_letters)
for l in ['R', 'M', 'A', 'K', 'C']:
    print(l, ":", counter[l])

输出：

R : 4
M : 3
A : 7
K : 3
C : 2

Answer 3

这是一种方法：

names = {"Renata", "Marek", "Maciek", "Karolina", "Marcel"}

# Create a dictionary that maps each letter that we're tracking to a frequency counter.
letter_counters = {'R': 0, 'M': 0, 'A': 0, 'K': 0, 'C': 0}

# Go through the set of names, one at a time, and add the frequency of each letter that we're tracking to our counters.
# .upper() is used to make our counting case-insensitive.
for name in names:
    for letter in letter_counters:
        letter_counters[letter] += name.upper().count(letter)

# Transform the dictionary into a list. (This could be made more concise using a list comprehension, though this is not immediately relevant to the question.)
result = []
for letter in letter_counters:
   result.append([letter, letter_counters[letter])