-1
defaultdict(<type 'int'>, {'match': 1})
[(0, 0)]
[]
[]
[]
Traceback (most recent call last):
  File "Joinomattic_1.py", line 409, in <module>
    verboseprint (magic(matching))
  File "Joinomattic_1.py", line 408, in <lambda>
    magic = lambda matching: int(''.join(str(i) for i in matching)) # Generator exp.
ValueError: invalid literal for int() with base 10: '''

上面的错误在我下面的代码中不断发生,但是它仅在较大程序的一部分时发生,当它自己运行时不会发生错误并且输出被写入 output_file。这个错误是什么意思,为什么它只发生在其他代码中?以下是我作为大型程序一部分的代码:

n = 0
consec_matches = []
chars = defaultdict(int)

for k, group in groupby(zip(line1_u_i, line2_u_rev_comp_i), class_chars):
    elems = len(list(group))
    chars[k] += elems
    if k == 'match':
        consec_matches.append((n, n+elems-1))
    n += elems

verboseprint (chars)
verboseprint (consec_matches)
verboseprint ([x for x in consec_matches if x[1]-x[0] >= 9])
list = [x for x in consec_matches if x[1]-x[0] >= 9]
flatten_list= [x for y in list for x in y]
verboseprint (flatten_list)
matching=[y[1] for y in list for x in y if x ==0 ]
verboseprint (matching)
magic = lambda matching: int(''.join(str(i) for i in matching)) # Generator exp.
verboseprint (magic(matching))
line2_u_rev_comp_i_l = line2_u_rev_comp_i[magic(matching):]
line3=line1_u_i+line2_u_rev_comp_i_l
verboseprint (line3)
if line3:
    verboseprint(line3, file = output_file)

在这里它是独立的:

from __future__ import print_function
from collections import defaultdict
from itertools import groupby
import argparse #imports the argparse module so it can be used
from itertools import izip





parser = argparse.ArgumentParser() #simplifys the wording of using argparse as stated in the python tutorial
parser.add_argument("-r1", type=str, action='store',  dest='input1', help="input the forward read file") # allows input of the forward read
parser.add_argument("-r2", type=str, action='store', dest='input2', help="input the reverse read file") # allows input of the reverse read
parser.add_argument("-v", "--verbose", action="store_true", help=" Increases the output, only needs to be used to provide feedback to Tom for debugging")
parser.add_argument("-n", action="count", default=0, help="Allows for up to 5 mismatches, however this will reduce accuracy of matching and cause mismatches. Default is 0")
parser.add_argument("-fastq", action="store_true", help=" States your input as fastq format")
parser.add_argument("-fasta", action="store_true", help=" States your input as fasta format")
parser.add_argument("-o", "--output", help="Directs the output to a name of your choice")
args = parser.parse_args()

output = str(args.output)


def class_chars(chrs):
    if 'N' in chrs:
        return 'unknown'
    elif chrs[0] == chrs[1]:
        return 'match'
    else:
        return 'not_match'


#with open(args.output, 'w') as output_file:

output_file= open(output, "w")

s1 = 'aaaaaaaaaaN123bbbbbbbbbbQccc'
s2 = 'aaaaaaaaaaN456bbbbbbbbbbPccc'
n = 0
consec_matches = []
chars = defaultdict(int)

for k, group in groupby(zip(s1, s2), class_chars):
    elems = len(list(group))
    chars[k] += elems
    if k == 'match':
        consec_matches.append((n, n+elems-1))
    n += elems

print (chars)
print (consec_matches)
print ([x for x in consec_matches if x[1]-x[0] >= 9])
list = [x for x in consec_matches if x[1]-x[0] >= 9]
flatten_list= [x for y in list for x in y]
print (flatten_list)
matching=[y[1] for y in list for x in y if x ==0 ]
print (matching)
magic = lambda matching: int(''.join(str(i) for i in matching)) # Generator exp.
print (magic(matching))
s2_l = s2[magic(matching):]
line3=s1+s2_l
print (line3)
if line3:
    print(line3, file = output_file)

独立时输出为:

defaultdict(<type 'int'>, {'not_match': 4, 'unknown': 1, 'match': 23})
[(0, 9), (14, 23), (25, 27)]
[(0, 9), (14, 23)]
[0, 9, 14, 23]
[9]
9
aaaaaaaaaaN123bbbbbbbbbbQcccaN456bbbbbbbbbbPccc
4

1 回答 1

0

正如您的输出所示,matching是一个空列表。因此,在 之后join,您会得到一个空字符串'',它不能转换为整数。尝试:

lambda matching: int(''.join(str(i) for i in matching) or 0)

return 0如果matching == []. _

于 2014-05-09T16:07:54.087 回答