Nested loop on list string python ,-CodePudding

I want to iterate the second loop through the second position

My error is

Exception has occurred: TypeError (note: full exception trace is shown but execution is paused at: ) can only concatenate list (not "int") to list

s = "dog cat cat dog"

b = list(s.split(" "))

for x in b:

    for y in str(b 1):

        print(x,'/', y)

CodePudding user response：

A way to fix your code is:

s = "dog cat cat dog"

b = list(s.split(" "))

for x in b:

    for y in b[1:]:

        print(x,'/', y)

Even though I don't think it does what it is supposed to do, because the output is:

dog / cat
dog / cat
dog / dog
cat / cat
cat / cat
cat / dog
cat / cat
cat / cat
cat / dog
dog / cat
dog / cat
dog / dog

Maybe you wanted something like

s = "dog cat cat dog"
b = list(s.split(" "))
for x,y in zip(b,b[1:]):
    print(x,'/', y)

Output:

dog / cat
cat / cat
cat / dog

CodePudding user response：

Answer based on your comment

If you simply want to loop through a list twice where the second loop is the remainder of the list, you can use

s = "dog cat cat dog"
b = s.split(" ")

for i in range(len(b)):
    for j in range(i 1, len(b)):
        # i is the first index, j is the indices of the remaining entries
        # So x = b[i], y = b[j]

Old answer

I am not sure I fully understand what you're searching for, perhaps something like this?

s = "dog cat cat dog"

b = s.split(" ")

for i in range(len(b)):
    for j in range(i 1, len(b)):
        print(b[i],'/', b[j])

which returns

dog / cat
dog / cat
dog / dog
cat / cat
cat / dog
cat / dog

Also, you don't need to cast a call to split() to a list, as it already returns a list.