如何打印出每个lin的元组元素

Question

下面是我的DNA串相邻问题的代码：

chars = "ACGT"

def neighbors(pattern, d):
    assert(d <= len(pattern))

    if d == 0:
        return [pattern]

    r2 = neighbors(pattern[1:], d-1)
    r = [c + r3 for r3 in r2 for c in chars if c != pattern[0]]

    if (d < len(pattern)):
        r2 = neighbors(pattern[1:], d)
        r += [pattern[0] + r3 for r3 in r2]

    return r
def neighbors2(pattern, d):
    return ([neighbors(pattern, d2) for d2 in range(d + 1)], [])

print (neighbors2("ACG", 1))

输出如下：

^{pr2}$

如何添加一些代码并将输出更改为这种模式：

CCG
TCG
GCG
AAG
ATG
AGG
ACA
ACC
ACT
ACG

Answer 1

您可以从编译器.ast模块

from compiler.ast import flatten
print flatten(neighbors2("ACG", 1))

会产生

^{pr2}$

或者

print("\n".join(flatten(neighbors2("ACG", 1))))

要获得这样的输出：

ACG
CCG
GCG
TCG
AAG
AGG
ATG
ACA
ACC
ACT

Answer 2

有几种方法可以做到这一点，你可以打印每一个，制作一个大字符串并打印它，使用print函数的sep参数，使一个表示你的东西的类成为一个定义的__str__方法，该方法返回你想要的字符串。在

例如

>>> test=['CCG', 'GCG', 'TCG', 'AAG', 'AGG', 'ATG', 'ACA', 'ACC', 'ACT']

打印每个

^{pr2}$

做一根大绳子

>>> print( "\n".join(test) )
CCG
GCG
TCG
AAG
AGG
ATG
ACA
ACC
ACT

使用sep和拆包

>>> print( *test, sep="\n" )
CCG
GCG
TCG
AAG
AGG
ATG
ACA
ACC
ACT

使用类

>>> class Foo:
    def __init__(self,data):
        self.data=data
    def __str__(self):
        return "\n".join(self.data)


>>> x=Foo(test)
>>> print(x)
CCG
GCG
TCG
AAG
AGG
ATG
ACA
ACC
ACT

为了从([['ACG'], ['CCG', 'GCG', 'TCG', 'AAG', 'AGG', 'ATG', 'ACA', 'ACC', 'ACT']], [])到{}，你可以使用这个Flatten (an irregular) list of lists in Python的答案，例如，unutbu的答案是我最喜欢的

from itertools import chain
from collections import Iterable

try: #python 2
    _basestring = basestring
except NameError:
    #python 3
    _basestring = (str,bytes)

def flatten_total(iterable, flattype=Iterable, ignoretype=_basestring):
    """Flatten all level of nesting of a arbitrary iterable"""
    #https://stackoverflow.com/questions/2158395/flatten-an-irregular-list-of-lists-in-python
    #unutbu version
    remanente = iter(iterable)
    while True:
        elem = next(remanente)
        if isinstance(elem,flattype) and not isinstance(elem,ignoretype):
            remanente = chain( elem, remanente )
        else:
            yield elem

然后去做

print( "\n".join(flatten_total( neighbors2("ACG", 1))) )

Answer 3

如果您只想打印并且嵌套级别始终相同，可以执行以下操作：

for entry in neighbors2("ACG", 1)[0]:
    print(*entry, sep='\n')

输出：

^{pr2}$