如何在python中统计给定名称(字符串)中每个字符的出现次数，并在不使用dict的情况下以正常格式显示？

Question

假设我的输入是：

marcel bentok tanaka

输出应为

m-1,a-4,r-1,c-1,l-1,e-2,l-1,b-1,n-2,t-2,o-1,k-2

我不想计算空格，也不想使用dict或导入functions.How，我能以最简单的方式实现这一点吗？请帮帮忙。我试着在网上搜索，但它们都以dict格式显示输出，即{m-1,a-4,r-1,c-1,l-1,e-2,l-1,b-1,n-2,t-2,o-1,k-2}。我刚刚开始学习编码。所以请帮帮忙。提前谢谢。

我试过了

def letter_count(name)
for char in name:
        print(f"{char}-{name.count(char)}")

但是这对输入bob不起作用。

Answer 1

仍然使用字典，但以更简单的方式，这是O(n^2)

def count_freq(name):
    x = dict()

    name = name.replace(' ', '')

    for i in name:
        if i in x.keys():
            x[i] += 1
        else:
            x[i] = 1

    b = [f"{k}-{v}" for k, v in x.items()]
    print(",".join(b))

name = input("Please insert your input: ")
count_freq(name)

这是一个初学者风格的O(n)解决方案：

def count_freq(name):
    x = dict()
    name = name.replace(' ', '')

    for i in name:
        try:
            x[i] += 1
        except:
            x[i] = 1

    b = [f"{k}-{v}" for k, v in x.items()]
    print(",".join(b))

name = input("Please insert your input: ")
count_freq(name)

执行时间分析：

#input is 'marcel bentok tanaka'
#string format no-print
theo 9 µs ± 1.37 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
**vn2  4.39 µs ± 157 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)**
vn   6.78 µs ± 1.34 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#no string format (counts only)
theo 5.74 µs ± 953 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
**vn2  2.42 µs ± 30 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)**
vn   3.66 µs ± 668 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#input is a really long text
#string format no-print
theo 20.8 µs ± 1.43 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
vn2  106 µs ± 21.9 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
**vn   6.06 µs ± 1.04 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)**

#no string format (counts only)
theo 16.6 µs ± 576 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
vn2  83 µs ± 21.2 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
**vn   3.22 µs ± 197 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)**

Answer 2

您可以尝试这样做：

def count_freq(name):
    name = name.replace(' ','')
    output = ''
    for char in set(name):
        output += (char + '-' + str(name.count(char)) + ',')
    return output

name = input('insert name:')
print(count_freq(name))

Answer 3

将其转换为字典，然后以任何您想要的方式将字典转换为字符串。例如,

counts = {}
for char in "abcdefghijklmnopqrstuvwxyz":
    counts[char] = myString.count(char)
for key in counts.keys():
    print(f"{key}-{counts[char]}")

或者，如果你真的讨厌口头语，那就跳过它：

for char in "abcdefghijklmnopqrstuvwxyz":
    print(f"{char}-{myString.count(char)}")

如果不想显示出现次数为零的内容，请将dict解决方案的最后一部分替换为：

for key in counts.keys():
    if counts[char]:
        print(f"{key}-{counts[char]}")