比较嵌套列表推导中的数字

Question

Can some one help me to compare numbers in nested list l=[[6, 6], [15, 24], [85, 18]] ; for example:

for i in l:
    if i[0][0]>i[0][1]:
        print("B")
    elif i[0][0]<i[0][1]:
        print("A")
    else:
        print("T")



Expected Output :
  T
  A
  B

  
    

Answer 1

You can try this:

l=[[6, 6], [23, 24], [85, 18]]

print(*['B' if i[0]>i[1] else ('A' if i[0]<i[1] else 'T')  for i in l], sep = "\n") 

Or if you want to use your original solution, just erase the i[0][0] because with the loop you are accesing to each list, so i is each nested list, eg [6, 6], [15, 24], [85, 18]

l=[[6, 6], [15, 24], [85, 18]]
for i in l:
    if i[0]>i[1]:
        print("B")
    elif i[0]<i[1]:
        print("A")
    else:
        print("T")

Both outputs:

T
A
B
    

Answer 2

Use meaningful variables to make your code more readable

for subarray in l:
    a,b = subarray
    if a > b:
        print("A")
    elif a < b:
        print("B")
    else:
        print("T")

Answer 3

You are accessing an array after entering the outer for loop so use index numbers as you would in a one dimensional array.

for i in l:
  if i[0] > i[1]:
    print("B")
  elif i[0] < i[1]
    print("A")
  else
    print("T")

Answer 4

You can :

l = [[6, 6], [15, 24], [85, 18]]
out = ["B" if x > y else ("A" if x < y else "T") for x, y in l]
print(out)
# ['T', 'A', 'B']

Would I use it? Not so sure but it really depends on your requirements.

Answer 5

your_list = [[6, 6], [15, 24], [85, 18]]

def compare(ab):
    a,b = ab
    if a > b: return "B"
    if a < b: return "A"
    if a == b: return "T"
        
res = "\n".join(map(compare, your_list))
print(res)