如何计算字符串中字母之间的平均位数？

Question

我有这样的字符串：

F-F-F+F+F+F-F-F-F+F+F+F-F+F+F-F+F-F-F-F

我想计算“ F”，“ +”和“-”之间的平均位数。
因此，对于此示例，它将是：

Average chars between Fs:   1
Average chars between +s:   2.25
Average chars between -s:   3

最有效的方法是什么？

Answer 1

这是一个变体。

首先我收集索引i所有出现的char acters; 然后我计算差异的mean ：

from collections import defaultdict
from itertools import islice
from statistics import mean

strg = "F-F-F+F+F+F-F-F-F+F+F+F-F+F+F-F+F-F-F-F"

dct = defaultdict(list)

for i, char in enumerate(strg):
    dct[char].append(i)

for char, occurrences in dct.items():
    avg = mean(b - a for a, b in zip(occurrences, islice(occurrences, 1, None))) - 1
    print(f"Average chars between {char}s:  {avg}")

打印：

Average chars between Fs:  1
Average chars between -s:  3
Average chars between +s:  2.25

在第一个for循环之后，在dct中将有这样的条目：

'-': [1, 3, 11, 13, 15, 23, 29, 33, 35, 37]

如前所述，第二个for循环计算差异的平均值。

Answer 2

我会通过以下方式使用正则表达式（ re模块）来做到这一点：

import re
txt = "F-F-F+F+F+F-F-F-F+F+F+F-F+F+F-F+F-F-F-F"
chars = set(list(txt))
between = dict()
for i in chars:
    between[i] = re.findall('(?<='+re.escape(i)+').*?(?='+re.escape(i)+')',txt)
for i in chars:
    if len(between[i])==0:
        between[i] = 0.0
    else:
        between[i] = sum([len(i) for i in between[i]])/len(between[i])
print(between)

输出：

{'F': 1.0, '+': 2.25, '-': 3.0}

说明：我正在以非贪婪的方式从左到右查找给定字符（使用零长度断言）的出现之间的子字符串（因此， "FFF"给出["-","-"]而不是["-F-"] ），然后简单地计算其长度的平均值。 请注意，我使用re.escape处理具有特殊含义的字符（例如+ ）。

Answer 3

不使用任何库的另一种方法：

string = 'F-F-F+F+F+F-F-F-F+F+F+F-F+F+F-F+F-F-F-F'
string = list(string)
chars = set(string)
for char in chars:
    ind = [i for i, x in enumerate(string) if x == char]
    diff = [ind[i+1]-ind[i] - 1 for i in range(len(ind)-1)]
    print(f'Average chars between {char}s:  {sum(diff) / len(diff)}')

输出：

Average chars between +s:  2.25
Average chars between Fs:  1.0
Average chars between -s:  3.0