我正在尝试解决 Rosalind 上的“共识和简介”挑战。挑战说明如下:
给定:最多 10 个等长(最多 1 kbp)的 FASTA 格式 DNA 字符串的集合。
返回:集合的共识字符串和配置文件矩阵。(如果存在多个可能的共识字符串,那么您可以返回其中任何一个。)
我的代码如下(我从这个网站上的另一个用户那里得到了大部分)。我唯一的问题是一些 DNA 链被分解成多个单独的行,因此它们被作为单独的字符串附加到“allstrings”列表中。我试图弄清楚如何将不包含“>”的每个连续行写为单个字符串。
import numpy as np
seq = []
allstrings = []
temp_seq = []
matrix = []
C = []
G = []
T = []
A = []
P = []
consensus = []
position = 1
file = open("C:/Users/knigh/Documents/rosalind_cons (3).txt", "r")
conout = open("C:/Users/knigh/Documents/consensus.txt", "w")
# Right now, this is reading and writing each as an individual line. Thus, it
# is splitting each sequence into multiple small sequences. You need to figure
# out how to read this in FASTA format to prevent this from occurring
desc = file.readlines()
for line in desc:
allstrings.append(line)
for string in range(1, len(allstrings)):
if ">" not in allstrings[string]:
temp_seq.append(allstrings[string])
else:
seq.insert(position, temp_seq[0])
temp_seq = []
position += 1
# This last insertion into the sequence must be performed after the loop to empty
# out the last remaining string from temp_seq
seq.insert(position, temp_seq[0])
for base in seq:
matrix.append([pos for pos in base])
M = np.array(matrix).reshape(len(seq), len(seq[0]))
for base in range(len(seq[0])):
A_count = 0
C_count = 0
G_count = 0
T_count = 0
for pos in M[:, base]:
if pos == "A":
A_count += 1
elif pos == "C":
C_count += 1
elif pos == "G":
G_count += 1
elif pos == "T":
T_count += 1
A.append(A_count)
C.append(C_count)
G.append(G_count)
T.append(T_count)
profile_matrix = {"A": A, "C": C, "G": G, "T": T}
P.append(A)
P.append(C)
P.append(G)
P.append(T)
profile = np.array(P).reshape(4, len(A))
for pos in range(len(A)):
if max(profile[:, pos]) == profile[0, pos]:
consensus.append("A")
elif max(profile[:, pos]) == profile[1, pos]:
consensus.append("C")
elif max(profile[:, pos]) == profile[2, pos]:
consensus.append("G")
elif max(profile[:, pos]) == profile[3, pos]:
consensus.append("T")
conout.write("".join(consensus) + "\n")
for k, v in profile_matrix.items():
conout.write(k + ": " + " ".join(str(x) for x in v) + "\n")
conout.close()