Python简介-列出问题

Question

we've started doing Lists in our class and I'm a bit confused thus coming here since previous questions/answers have helped me in the past.

The first question was to sum up all negative numbers in a list, I think I got it right but just want to double check.

import random

def sumNegative(lst):
    sum = 0
    for e in lst:
        if e < 0:
            sum = sum + e
    return sum

lst = []
for i in range(100):
    lst.append(random.randrange(-1000, 1000))

print(sumNegative(lst))

For the 2nd question, I'm a bit stuck on how to write it. The question was: Count how many words occur in a list up to and including the first occurrence of the word “sap”. I'm assuming it's a random list but wasn't given much info so just going off that.

I know the ending would be similar but no idea how the initial part would be since it's string opposed to numbers.

I wrote a code for a in-class problem which was to count how many odd numbers are on a list(It was random list here, so assuming it's random for that question as well) and got:

import random

def countOdd(lst):
    odd = 0
    for e in lst:
        if e % 2 = 0:
            odd = odd + 1
    return odd

lst = []
for i in range(100):
    lst.append(random.randint(0, 1000))

print(countOdd(lst))

How exactly would I change this to fit the criteria for the 2nd question? I'm just confused on that part. Thanks.

Answer 1

The code to sum -ve numbers looks fine! I might suggest testing it on a list that you can manually check, such as:

print(sumNegative([1, -1, -2]))

The same logic would apply to your random list.

A note about your countOdd function, it appears that you are missing an = ( == checks for equality, = is for assignment) and the code seems to count even numbers, not odd. The code should be:

def countOdd(lst):
    odd = 0
    for e in lst:
        if e%2 == 1:       # Odd%2 == 1
            odd = odd + 1
    return odd

As for your second question, you can use a very similar function:

def countWordsBeforeSap(inputList):
    numWords = 0
    for word in inputList:
        if word.lower() != "sap":      
            numWords = numWords + 1
        else:
            return numWords

inputList = ["trees", "produce", "sap"]
print(countWordsBeforeSap(inputList))

To explain the above, the countWordsBeforeSap function:

• Starts iterating through the words.
• If the word is anything other than "sap" it increments the counter and continues
• If the word IS "sap" then it returns early from the function

The function could be more general by passing in the word that you wanted to check for:

def countWordsBefore(inputList, wordToCheckFor):
    numWords = 0
    for word in inputList:
        if word.lower() != wordToCheckFor:      
            numWords = numWords + 1
        else:
            return numWords

inputList = ["trees", "produce", "sap"]
print(countWordsBeforeSap(inputList, "sap"))

If the words that you are checking come from a single string then you would initially need to split the string into individual words like so:

inputString = "Trees produce sap"
inputList   = inputString.split(" ")

Which splits the initial string into words that are separated by spaces.

Hope this helps! Tom

Answer 2

def count_words(lst, end="sap"):
    """Note that I added an extra input parameter.
       This input parameter has a default value of "sap" which is the actual question.
       However you can change this input parameter to any other word if you want to by
       just doing "count_words(lst, "another_word".
    """
    words = []
    # First we need to loop through each item in the list.
    for item in lst:
        # We append the item to our "words" list first thing in this loop,
        # as this will make sure we will count up to and INCLUDING.
        words.append(item)

        # Now check if we have reached the 'end' word.
        if item == end:
            # Break out of the loop prematurely, as we have reached the end.
            break
    # Our 'words' list now has all the words up to and including the 'end' variable.
    # 'len' will return how many items there are in the list.
    return len(words)

lst = ["something", "another", "woo", "sap", "this_wont_be_counted"]
print(count_words(lst))

Hope this helps you understand lists better!

Answer 3

You can make effective use of list/generator comprehensions. Below are fast and memory efficient.

1. Sum of negatives:

print(sum( i<0 for i in lst))

2. Count of words before sap: Like you sample list, it assumes no numbers are there in list.

print(lst.index('sap'))

If it's a random list. Filter strings. Find Index for sap

l = ['a','b',1,2,'sap',3,'d']
l = filter(lambda x: type(x)==str, l)
print(l.index('sap'))

3. Count of odd numbers:

print(sum(i%2 != 0 for i in lst))