Question

new_list=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"] 

all_possible=[] #print of all possible combinations
y=0
z=0
def writer():
    for z in range(0, 26):
        all_possible.append(new_list[0] + "" + new_list[y] + "" + new_list[z])
    z = 0

def reader():
    for y in range(0, 26):
        writer()

reader()
print(all_possible)  
all_possible=[]



for y in range(0, 26):
    for z in range(0, 26):
        all_possible.append(new_list[0] + "" + new_list[y] + "" +new_list[z])
    z = 0
print(all_possible)

所以我创建了这段代码以打印一个列表，该列表打印出所有包含3个字符的字母组合，我也想使用一个函数来执行此操作，但是由于某种原因，当我将相同的内容放入函数中时，它并没有由于某些原因，以相同的方式打印。你能帮忙吗？

Answer 1

您的函数应接受参数，以便您可以传递列表引用以修改函数：

new_list=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

all_possible=[] #print of all possible combinations
def writer(l, y):
    for z in range(0, 26):
        l.append(new_list[0] + "" + new_list[y] + "" + new_list[z])

def reader(l):
    for y in range(0, 26):
        writer(l, y)

reader(all_possible)
print(all_possible)

Answer 2

如果您希望两个列表都提供相同的输出，则应向函数编写器添加一个参数。否则，y的值将始终为零，并且永远不变。

new_list=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"] 
all_possible1=[] #print of all possible combinations
y=0
z=0
def writer(y):
    for z in range(0, 26):
        all_possible1.append(new_list[0] + "" + new_list[y] + "" + new_list[z])
    z = 0

def reader():
    for y in range(0, 26):
        writer(y)

reader()
print(all_possible1)
all_possible2=[]



for y in range(0, 26):
    for z in range(0, 26):
        all_possible2.append(new_list[0] + "" + new_list[y] + "" +new_list[z])
    z = 0
print(all_possible2)
print(all_possible2 == all_possible1)

Answer 3

如果您想轻松做到

import itertools
list(itertools.combinations(new_list, 3))

但是根据您发布的代码，我认为您将希望按照自己的方式做，并了解有关python语法的更多信息，因此此答案仅是一个建议

为什么不打印相同的列表

3 个答案: