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3个字符串的列表（seq_1-3）。我将它们组合起来创建一个长串（组合）我有一个索引列表（联结）。我为每个字符串设置了3个不同的计数器（count_1-3）

我想要做的是找到组合序列中每个交叉点[2,9,15,20]的位置。。。如果是来自seq_1 - ＆gt; count_1 + = 1，如果是来自seq_2 - ＆gt; count_2 + = 1，来自seq_3 - ＆gt; count_3 + = 1

例如

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

让我知道我是否需要澄清任何不同的

Answer 1

您可以在此处使用collections.Counter和bisect.bisect_left：

>>> from collections import Counter
>>> import bisect
>>> junctions = [2,9,15,20]
>>> seq_1 = 'sauron'
>>> seq_2 = 'corrupted'
>>> seq_3 = 'numenor'
>>> lis  = [seq_1, seq_2, seq_3]

创建一个列表，其中包含每个seq_结束的索引：

>>> start = -1
>>> break_points = []
for item in lis:
    start += len(item) 
    break_points.append(start)
...     
>>> break_points
[5, 14, 21]

现在我们可以简单地循环junctions并使用break_points函数找到bisect.bisect_left列表中每个联结的位置。

>>> Counter(bisect.bisect_left(break_points, jun)+1  for jun in junctions)
Counter({3: 2, 1: 1, 2: 1})

使用collections.defaultdict更好地输出：

>>> from collections import defaultdict
>>> dic = defaultdict(int)
for junc in junctions:
    ind = bisect.bisect_left(break_points, junc) +1
    dic['count_'+str(ind)] += 1
...     
>>> dic
defaultdict(<type 'int'>,
{'count_3': 2,
 'count_2': 1,
 'count_1': 1})

#accessing these counts
>>> dic['count_3']
2

Answer 2

你可以尝试一些基本的东西，比如

L_1 = len(seq_1)
L_2 = len(seq_2)
L_3 = len(seq_3)

junctions = [2, 9, 15, 20]
c_1, c_2, c_3 = (0, 0, 0)

for j in junctions:
    if j < L_1:
        c_1 += 1
    elif j < L_1 + L_2:
        c_2 += 1
    elif j < L_1 + L_2 + L_3:
        c_3 += 1
    else:
        Raise error

Answer 3

可以使用来自itertools的collections.Counter，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})

结合字符串。计算原始字符串中有多少个索引（来自列表）。蟒蛇

3 个答案: