Question

我有这个字符串

str = CTGGCATAACAAGACAAAAACAAAAGCAATAAATGCTGAAAAAACAAAATGCCGTGATCGTTTGTAATACTGGAACATAGTCATGATGAATGAAGGTTTCTGAACCTGAAGAACGACCTGAAAAAGTCAAACCGCAAGAATATCACGACGCAGTGAACCAGAATAGCAACGACGAAAATGTCCAGGAAAAATCCTGGAGTCAGATTCAGGGTTATTCGTTAGTGGCAGGATTACGAAGCGTGGGGCACAGGAGATACATCTCCAGTAAGATGGCAACGTAATCGCGGGCTTCTTTTTTAAGATCAAAAGATTGCGGGGCAAAGAGCCAGTTTTCCATCAGGCCGGAAATATAGCCGCGCATAATAATTGCTGCGCGACGCGTCATTAAATCCGCAGGCAACATTTTCGCTTCAATACAATGTTTTAACGTTTGTTCTATACGGTCATAACTTTCCAGACAGAGATTACGTTGTGCCTGTTGCACAACAGCCATTTCTCCGACAAATTCGCATTTGTGGAATATAATCTCCATCAATAATCGACGCCGTTCTTCTGTCACCGTGGATTCAAGAACATGAATTAATATCTCTCTTAATACTGAGAGTGGATCGCCAGGGAATTTTGCCTGATACTCAAGCTCTAGTTCACCAATATTGGATTCTGACAGTTCCCAGATCTCACTGAACAAATCCGACTTGTCTTTAAAATGCCAGTAGATTGCACCGCGCGTAACGCCAGCTGCTTTTGCAATCTCGCCCAGCGAGGTGGATGATACCCCCTGCTGTGAGAAAAGACGTAGAGCCACATCGAGGATGTGTTGGCGCGTTTCTTGCGCTTCTTGTTTGGTTTTTCGTGCCATATGTTCGTGAATTTACAGGCGTTAGATTTACATACATTTGTGAATGTATGTACCATAGCACGACGATAATATAAACGCAGCAATGGGTTTATTAACTTTTGACCATTGACCAATTTGAAATCGGACACTCGAGGTTTACATA

我想在str [0:19] str [1:20] str [2:21] str [3:22] .....等多个子字符串中剪切这个字符串到最后。

Answer 1

使用字符串切片：

>>> strs = "CTGGCATAACAAGACAAAAACAAAAGCAATAAATGCTGAAAAAACAAAATGCCGTGATCGTTTGTAATACTGGAACATAGTCATGATGAATGAAGGTTTCTGAACCTGAAGAACGACCTGAAAAAGTCAAACCGCAAGAATATCACGACGCAGTGAACCAGAATAGCAACGACGAAAATGTCCAGGAAAAATCCTGGAGTCAGATTCAGGGTTATTCGTTAGTGGCAGGATTACGAAGCGTGGGGCACAGGAGATACATCTCCAGTAAGATGGCAACGTAATCGCGGGCTTCTTTTTTAAGATCAAAAGATTGCGGGGCAAAGAGCCAGTTTTCCATCAGGCCGGAAATATAGCCGCGCATAATAATTGCTGCGCGACGCGTCATTAAATCCGCAGGCAACATTTTCGCTTCAATACAATGTTTTAACGTTTGTTCTATACGGTCATAACTTTCCAGACAGAGATTACGTTGTGCCTGTTGCACAACAGCCATTTCTCCGACAAATTCGCATTTGTGGAATATAATCTCCATCAATAATCGACGCCGTTCTTCTGTCACCGTGGATTCAAGAACATGAATTAATATCTCTCTTAATACTGAGAGTGGATCGCCAGGGAATTTTGCCTGATACTCAAGCTCTAGTTCACCAATATTGGATTCTGACAGTTCCCAGATCTCACTGAACAAATCCGACTTGTCTTTAAAATGCCAGTAGATTGCACCGCGCGTAACGCCAGCTGCTTTTGCAATCTCGCCCAGCGAGGTGGATGATACCCCCTGCTGTGAGAAAAGACGTAGAGCCACATCGAGGATGTGTTGGCGCGTTTCTTGCGCTTCTTGTTTGGTTTTTCGTGCCATATGTTCGTGAATTTACAGGCGTTAGATTTACATACATTTGTGAATGTATGTACCATAGCACGACGATAATATAAACGCAGCAATGGGTTTATTAACTTTTGACCATTGACCAATTTGAAATCGGACACTCGAGGTTTACATA"
>>> substrings = [strs[i:i+19] for i in xrange(len(strs))]
>>> substrings
['CTGGCATAACAAGACAAAA', 'TGGCATAACAAGACAAAAA', 'GGCATAACAAGACAAAAAC',...]

Answer 2

chopped_str = []
for i in range(0, len(str)-19):
   chopped_str.append(str[i:i+19])

Answer 3

如果你想从链中提取19个核苷酸的所有序列，那么就可以了：

>>> SEQ_LEN = 20
>>> [strs[i:i+SEQ_LEN] for i in xrange(len(strs)-SEQ_LEN)]

但是，它的内存效率不高，因为它会产生所有子序列的列表。它的用途是什么？

处理N个核苷酸的每个亚序列的另一种方法可以是：

for seq in (strs[i:i+SEQ_LEN] for i in xrange(len(strs)-SEQ_LEN)):
    do_something_with(seq)

针对您的具体问题，do_something_with将主要使用核苷酸位置更新PWM。如果您遇到困难，请随意发布其他问题;）

python中字符串的子串

3 个答案: