Sure! Here is the same script that  an find a specific member in the sorted list:

```python

# the list to be sorted
l = [1000, 102, 110]

# search value
v = 110

# the sort() function takes an optional argument, the "key" which is used for sorting the elements in a list.
# this can be modified with a function. we can sum the ord values from each character (in this case digit) in x
# (the current number) by first converting it to a list, then applying the ord function to each element.
# this way the list will be sorted using the sum of ord values of each digit of all elements.
# we first define the function, then use it as the key in the sort() function.
def sum_of_ord_values(x):
    return sum(map(ord, list(str(x))))

# sort the list using the sum_of_ord_values function as the key
sorted_list = sorted(l, key = lambda x: sum_of_ord_values(x))

print(sorted_list)
# The ord() values of 1000 are
# ord(str(1)) + ord(str(0)) + ord(str(0)) + ord(str(0)) = 49 + 48 + 48 + 48 = 193

# The ord() values of 102 are
# ord(str(1)) + ord(str(0)) + ord(str(2)) = 49 + 48 + 50 = 147

# The ord() values of 110 are
# ord(str(1)) + ord(str(1)) + ord(str(0)) = 49 + 49 + 48 = 146

# therefore sorting the list is equal to sorting [193, 147, 146], the result of which is [146, 147, 193]
# therefore, the expected output is: [110, 102, 1000]

# to find the value x in the list, we can use the built-in method index, i.e., sorted_list.index(v).
# but since this is a sorted list, we can use binary search algorithms to do this more efficiently.
# values in the sorted list must be compared by the key by which they were sorted, i.e., the sum of ord values.
low = 0  # algorithm search lower bound
high = len(sorted_list) - 1  # algorithm search higher bound
idx = -1  # variable to hold the resulting index if found
v_ord_key = sum_of_ord_values(v)  # the ord key of the value to be searched
while low <= high:

    mid = (high + low) // 2

    mid_ord_key = sum_of_ord_values(sorted_list[mid])

    # If x is greater, ignore left half
    if mid_ord_key < v_ord_key:
        low = mid + 1

    # If x is smaller, ignore right half
    elif mid_ord_key > v_ord_key:
        high = mid - 1

    # means x is present at mid
    else:
        idx = mid
        break

# check if value was found in binary search
if idx >= 0:
    # value found!
    print(f'element “{v}” found at index {idx}')
else:
    # value not found
    print(f'element “{v}” was not found in the list')

# as previously mentioned, the sorted list is [110, 102, 1000].
# the search will begin with 102, whose ord value is 147, which is greater than 146, which is the ord value of 110.
# therefore, the search will continue with the left half of the list, i.e., [110]. the value will be found there.
# this is the first value in the sorted list, and thus the output index is 0.

```
spam fails_task lang_mismatch pii not_appropriate hate_speech sexual_content quality toxicity humor helpfulness creativity violence