Question

我在函数实际字符串中遇到了return语句的问题。当我单独运行时，该函数正常运行，但是当我在程序中运行时，它没有返回结果。函数real string应返回结果如下：((.((....)).).)

import numpy
import array
class Structure :
    def __init__(self): 
        self.paired =[]
        self.unpaired =[]
    def merge(self, s):
        for i in s.paired :
            self.paired.append(i)
        for i in s.unpaired :
            self.unpaired.append(i)
    def __str__(self):
        return "paired\n"+str(self.paired) +"\n unpaired \n"+str(self.unpaired)+"\n"
def l(ch) :

    if (ch=='A') :
        return 0
    elif (ch=='C'):
        return 1
    elif (ch=='U') :
        return 2
    elif (ch =='G') :
        return 3
    else :
        print "error"

def cost(i,j,M,w,seq,compare): #Defining the cost functions or can say recurrence formula
    j_unpaired = M[i,j-1]
    i_unpaired = M[i+1,j]
    paired = M[i+1,j-1] + w[l(seq[i]), l(seq[j])]

    return compare(j_unpaired,i_unpaired,paired)  #return comparison of the three relations

def fill(M, w, seq, compare): #function defining how to fill up the values 
    n = len(seq)
    for k in range(5,n):
        for i in range(0,n-k):
            j = k+i
            c = cost(i,j,M,w,seq,compare)
            M[i,j] = c

def trace(M, w, seq, compare, horizontal_arrow, vertical_arrow) :
    i = horizontal_arrow
    j = vertical_arrow
    struct = Structure()
    while (j != i-1):
        if M[i,j] == M[i,j-1] :
            struct.unpaired.append(j)
            j -= 1

        elif M[i,j] == M[i+1,j] :

            struct.unpaired.append(i)

            i += 1
        elif M[i,j] == M[i+1,j-1] + w[l(seq[i]), l(seq[j])] :

            struct.paired.append((i,j))
            i += 1
            j -= 1
        else:
            print "error"

    return struct

def real_string(M, w, seq, compare, horizontal_arrow, vertical_arrow) : 
    trace=dict()
    struct = Structure()
    for element in struct.unpaired:
        trace[element] = '.'

    for element in struct.paired:
        trace[element[0]] = '('
        trace[element[1]] = ')'

    return list(trace.values())##The glitch is here;not showing result.

f =open("example.fasta","r")
def readSeq(FASTA):
    for line in FASTA:
        if line.startswith('>'):
            continue
        line = line.strip()
        return line


seq=readSeq(f)

n = len(seq)

w = numpy.zeros([4,4],int)
w.fill(0)
w[l('A'), l('U')] = 1
w[l('U'), l('A')] = 1 
w[l('G'), l('C')] = 1
w[l('C'), l('G')] = 1
M = numpy.zeros([n,n],int) #initialisation of matrix
fill(M, w, seq, max) #Filling matrix M with scores
print M
# perform traceback 
s = trace(M, w, seq, max, 0, n-1)
print str(s)
print seq
result=real_string(M, w, seq, max,0,n-1)
print "".join(result)

当我运行代码时，它可以工作：

yourDict=dict()

unpaired = [13, 11, 2, 8, 7, 6, 5]

paired = [(0, 14), (1, 12), (3, 10), (4, 9)]



for element in unpaired:
    yourDict[element] = '.'

for element in paired:
    yourDict[element[0]] = '('
    yourDict[element[1]] = ')'

print "".join(list(yourDict.values()))

Answer 1

您的real_string函数初始化新的Structure：

struct = Structure()

__init__的{{1}}为配对和未配对创建空列表。然后，您在Structure中迭代这些空列表，这些列表不会将任何内容放入real_string字典中。

因此该函数返回一个空列表。

我不知道你究竟想要达到的目的，但似乎你没有使用传递给你的函数的大多数参数。很可能您想使用trace返回的s：

trace

所以将它作为s = trace(M, w, seq, max, 0, n-1)的参数并删除struct的初始化。

Answer 2

将real_string更改为此，因为它有错误的参数：

def real_string(struct) : 
    trace=dict()
    for element in struct.unpaired:
        trace[element] = '.'

    for element in struct.paired:
        trace[element[0]] = '('
        trace[element[1]] = ')'

    return list(trace.values())

并将其称为：

# perform traceback 
s = trace(M, w, seq, max, 0, n-1)
print str(s)
print seq
result=real_string(s)
print "".join(result)

return语句出错

2 个答案: