如何获取嵌套列表范围内的元素

Question

Just started learning python, I have this assignment where I need do range within nested list.

tlist = [['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4], ['Zen', 'No', 90, 1]]

I want to ask for the low and higher value for tlist[2]

low = int(input("Enter low")) #take this as 1
high = int(input("Enter High")) #take this as 50

this need to be updated and when I call the list again it should show, and remove 'zen' from the list

tlist = [['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4]]

Answer 1

You can use this code. Try to iterate over the list with condition including low and high

tlist = [['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4], ['Zen', 'No', 90, 1]]
low = int(input("Enter low")) #take this as 1
high = int(input("Enter High")) #take this as 50
tlist = [l for l in tlist if l[2]>=low and l[2]<=high]
print(tlist)

Answer 2

Method 1 : Keeping tlist unchanged and saving result to another list.

• Loop through each inner list in t_list
• Access the third element at index 2 in each inner list.
• Check if this element in the range [low,high] .
• If yes, add current inner list to new list.

tlist = [['Foo', 'No', 80, 1], ['Bar', 'No', 90, 1], ['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4], ['Zen', 'No', 90, 1]] 
updated_list = []
low = 1
high = 50

for inner_list in tlist:
    if inner_list[2] >= low and inner_list[2] <= high:
        updated_list.append(inner_list)

print(updated_list)
# [['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4]]

Method 2 : Using remove() to directly modify tlist

tlist = [['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4], ['Zen', 'No', 90, 1]]
low = 1
high = 50
for i in range(len(tlist)-1,-1,-1):
    inner_list = tlist[i]
    if inner_list[2] < low or inner_list[2] > high:
        tlist.remove(inner_list)
print(tlist)
# [['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4]]

When tlist = [['Foo', 'No', 80, 1], ['Bar', 'No', 90, 1], ['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4], ['Zen', 'No', 90, 1]] , output is :

[['Josh', 'Yes', 1, 4], ['Amy', 'No', 30, 4]]

Note : The concern expressed in the comments has been resolved by iterating from the end of the list instead of the beginning. The re-indexing caused by remove() no longer affects the final answer.