Python：列表中的最小数字

Question

I'm new to python and i don't know how to find the smallest number in each array. The input is [ [2], [3, 4], [6, 5, 7], [4, 1, 8, 3] ] and the output should be [2, 3, 5, 1] . I have try this:

x = [ [2], [3, 4], [6, 5, 7], [4, 1, 8, 3] ]
minimal = x[0]
  for i in range(len(x)):
      if (x[i]< minimal):
          minimal = x[i]
print(minimal)

And the output that i got is [2] , i have no idea about this. Please help me...

Answer 1

We can simply iterate through the nested list and append the minimum values for each of the list item.

smallest = []
for listItem in x:
    smallest.append(min(listItem))

Now, smallest list should be the expected output list.

Answer 2

First, we need to get individual lists from list x using for lst in x: . Then we find the minimum value of that list using min() function & append it to your minimal list using minimal.append(min(lst)) , then print the final minimal list Now you will have your output as [2, 3, 5, 1]

Now try & understand this code:

x = [ [2], [3, 4], [6, 5, 7], [4, 1, 8, 3] ]

minimal = []

for lst in x:
    minimal.append(min(lst))
print(minimal)

Answer 3

try this:

list(map(lambda x:min(x),[[2], [3, 4], [6, 5, 7], [4, 1, 8, 3] ]))