kdy00963990

<pre><code>A코드

func fetchAsyncLetTwoImages() async throws -> (UIImage, UIImage) {
    let start = Date()
    /// 구조적 동시성 작업의 생성 (하위 작업의 생성)
    async let image1 = try await fetchImage(num: 1)
    async let image2 = try await fetchImage(num: 2)
    
    let images = try await (image1, image2)
    print(Date().timeIntervalSince(start).formatted(.number.precision(.fractionLength(1))))
    return images
}

B코드
func fetchAsyncLetTwoImages() async throws -> (UIImage, UIImage) {
    let start = Date()
    /// 구조적 동시성 작업의 생성 (하위 작업의 생성)
    async let image1 = fetchImage(num: 1)
    async let image2 = fetchImage(num: 2)
    
    let images = try await (image1, image2)
    print(Date().timeIntervalSince(start).formatted(.number.precision(.fractionLength(1))))
    return images
}
</code></pre>

참고 코드 자료 7-StructuredConcurrency(102, 103줄) 오타?

<p>저는 \n, 컴마, 빈칸 으로 나눴고 \n 으로 나눈 경우 대부분 한 칸이 띄어져 있어서 단어의 리스트에 ""이 요소로 잡힙니다. 그래서 그냥 단어 리스트에서 ""를 제거하고 나니 73이 나오네요.</p><pre><code>import re

# 방법 1: regular expression 사용
def how_many_words_in_a_text_file(path: str) -> (str, int):
    with open(path, "r", encoding="utf-8") as f:
        text = f.read()
        words = re.split(" |,|\n", text)
        words = [word for word in words if word != ""]
        return (words, len(words))

print(how_many_words_in_a_text_file("../source/22-1.txt"))
---
결과값
(['The', 'adjective', '"deep"', 'in', 'deep', 'learning', 'refers', 'to', 'the', 'use', 'of', 'multiple', 'layers', 'in', 'the', 'network.', 'Early', 'work', 'showed', 'that', 'a', 'linear', 'perceptron', 'cannot', 'be', 'a', 'universal', 'classifier', 'but', 'that', 'a', 'network', 'with', 'a', 'nonpolynomial', 'activation', 'function', 'with', 'one', 'hidden', 'layer', 'of', 'unbounded', 'width', 'can.', 'Deep', 'learning', 'is', 'a', 'modern', 'variation', 'which', 'is', 'concerned', 'with', 'an', 'unbounded', 'number', 'of', 'layers', 'of', 'bounded', 'size', 'which', 'permits', 'practical', 'application', 'and', 'optimized', 'implementation', 'while', 'retaining', 'theoretical'], 73)

# 방법2 : 정규표현식 없이 replace와 split으로

def number_of_words_in_the_text(path: str) -> (str, int):
    with open(path, 'r', encoding="utf-8") as f:
        text = f.read()
        text = text.replace(",", " ").replace("\n", " ")
        words = text.split(" ")
        words = [word for word in words if word != ""]
        return (words, len(words))

print(number_of_words_in_the_text("../source/22-1.txt"))

--- 
결과값
(['The', 'adjective', '"deep"', 'in', 'deep', 'learning', 'refers', 'to', 'the', 'use', 'of', 'multiple', 'layers', 'in', 'the', 'network.', 'Early', 'work', 'showed', 'that', 'a', 'linear', 'perceptron', 'cannot', 'be', 'a', 'universal', 'classifier', 'but', 'that', 'a', 'network', 'with', 'a', 'nonpolynomial', 'activation', 'function', 'with', 'one', 'hidden', 'layer', 'of', 'unbounded', 'width', 'can.', 'Deep', 'learning', 'is', 'a', 'modern', 'variation', 'which', 'is', 'concerned', 'with', 'an', 'unbounded', 'number', 'of', 'layers', 'of', 'bounded', 'size', 'which', 'permits', 'practical', 'application', 'and', 'optimized', 'implementation', 'while', 'retaining', 'theoretical'], 73)</code></pre>

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