如何从 PYTHON I 中以“A”开头的列表中提取单词？

Question

The directions say "Use the following initializer list:

w = ["Algorithm", "Logic", "Filter", "Software", "Network", "Parameters", "Analyze", "Algorithm", "Functionality", "Viruses"] Write a loop to print the words that start with "A".

Sample Run Algorithm Analyze Algorithm"

I tried to do it with the following code:

w = ["Algorithm", "Logic", "Filter", "Software", "Network", "Parameters", "Analyze", "Algorithm", "Functionality", "Viruses"]

for i in range(len(w)):
     if (i[0] == "a" or i[0] == "A"):
          print(w[i])

but I keep getting the error "Line 4: TypeError: 'int' does not support indexing"

pls help

Answer 1

You can use a list comprehension for this

>>> [word for word in w if word[0].upper() == 'A']
['Algorithm', 'Analyze', 'Algorithm']

If you want to be case-specific (ie do not match both 'a' and 'A' ) you can remove the .upper()

Answer 2

You forgot to get the word out of the list and instead were trying to get the first character of the index (i), which obviously doesn't work.

w = ["Algorithm", "Logic", "Filter", "Software", "Network", "Parameters", "Analyze", "Algorithm", "Functionality", "Viruses"]

for i in range(len(w)):
     if (w[i][0] == "a" or w[i][0] == "A"):
          print(w[i])

However a more "pythonic" way of iterating over a list is as follows:

for word in w:
     if (word[0] == "a" or word[0] == "A"):
          print(word)

Answer 3

How do I extract words from a list that start with “A”

str.startswith does exactly that:

Return True if string starts with the prefix, otherwise return False. prefix can also be a tuple of prefixes to look for. With optional start, test string beginning at that position. With optional end, stop comparing string at that position.

>>> [word for word in w if word.startswith("A")]
['Algorithm', 'Analyze', 'Algorithm']

Answer 4

Another easier approach:

for word in w:
    if word.startswith("A"):
        print(word)