合并字符串。计算原始字符串中有多少个指数（来自列表）。Python

Question

junctions = [2,9,15,20]

seq_1 = 'sauron'
seq_2 = 'corrupted'
seq_3 = 'numenor'
combined = 'sauroncorruptednumenor' #seq_1 + seq_2 + seq_3

count_1 = 1
count_2 = 1
count_3 = 2

我有一个3个字符串的列表（序号1-3）。我将它们组合起来创建一个长字符串（组合） 我有一个索引（连接）列表。我为每个字符串设置了3个不同的计数器（计数1-3）

我要做的是找出组合序列中每个连接点[2,9,15,20]的位置。如果来自序列1-->；计数1+=1，如果来自序列2-->；计数2+=1，则来自序列3-->；计数3+=1

示例

junctions = [2,9,15,20]
count_1 = 0
count_2 = 0
count_3 = 0
combined = 'sauroncorruptednumenor'
seq_1 = 'sauron' #index 2 would be on 'u' in combined but originally from seq_1 so count_1 = count_1 + 1 
seq_2 = 'corrupted' #index 9 would be on 'r' in combined so count_2 += 1
seq_3 = 'numenor' #index 15 would be 'n' in combined so count_3 += 1, and 20 would be 'o' so count_3 += 1

如果我有什么需要澄清的，请告诉我

Answer 1

可以使用itertools中的collections.Counter、和repeat和chain，例如：

from itertools import chain, repeat
from operator import itemgetter
from collections import Counter

junctions = [2,9,15,20]
seq_1 = 'sauron'
seq_2 = 'corrupted'
seq_3 = 'numenor'

indices = list(chain.from_iterable(repeat(i, len(j)) for i, j in enumerate([seq_1, seq_2, seq_3], start=1)))
print Counter(itemgetter(*junctions)(indices))
# Counter({3: 2, 1: 1, 2: 1})