Question

我正在研究这个问题而找不到合适的答案，但我已经设法挖了一个更大的漏洞。迷惑自己。所以如果有人能提供清晰度：

方向：

编写一个函数analyze_text，它接收一个字符串作为输入。您的函数应该计算文本中字母字符的数量（a到z，或A到Z），并且还要跟踪字母的数量＆＃39; e＆＃39; （大写或小写）。

你的函数应该以字符串形式返回对文本的分析，其语法如下：

“该文字包含240个字母字符，其中105（43.75％）为'e'。”

您需要使用isalpha功能。

我的代码到目前为止： def analyze_text（text）： count = 0 letter_count = 0

for char in text:
    if char.isalpha():
        count += 1
for e in text:
    if e == "e" or e =="E":
        letter_count += 1
    p = float(letter_count)/float(count) * 100

analyze.text = "The text contains {0} alphabetic characters, of 
which {1} ({2}) are 'e'."
print(analyze_text.format(count += 1, letter_count += 1, p))

TESTS that are given:
# Note that depending on whether you use str.format or 
string concatenation

# your code will pass different tests. Code passes either
# tests 1-3 OR tests 4-6.

from test import testEqual

# Tests 1-3: solutions using string concatenation should pass these
text1 = "Eeeee"
answer1 = "The text contains 5 alphabetic characters, 
of which 5 (100.0%) are 'e'."
testEqual(analyze_text(text1), answer1)

text2 = "Blueberries are tasteee!"
answer2 = "The text contains 21 alphabetic characters, of 
which 7 (33.3333333333%) are 'e'."
testEqual(analyze_text(text2), answer2)

text3 = "Wright's book, Gadsby, contains a total of 0 of 
that most common symbol ;)"
answer3 = "The text contains 55 alphabetic characters, 
of which 0 (0.0%) are 'e'."
testEqual(analyze_text(text3), answer3)

# Tests 4-6: solutions using str.format should pass these
text4 = "Eeeee"
answer4 = "The text contains 5 alphabetic characters, 
of which 5 (100%) are 'e'."
testEqual(analyze_text(text4), answer4)

text5 = "Blueberries are tasteee!"
answer5 = "The text contains 21 alphabetic characters, 
of which 7 (33.33333333333333%) are 'e'."
testEqual(analyze_text(text5), answer5)

text6 = "Wright's book, Gadsby, contains a total of 
0 of that most common symbol ;)"
answer6 = "The text contains 55 alphabetic characters, 
of which 0 (0%)     are 'e'."
testEqual(analyze_text(text6), answer6)

Answer 1

评论中只列出了一些错误：

def analyze_text(text):
    count = 0
    letter_count = 0
    for char in text:
        if char.isalpha():
            count += 1
    for e in text:
        if e == "e" or e == "E":
            letter_count += 1
    p = float(letter_count) / float(count) * 100

    # here, you put a period instead of an underscore in analyze_text
    # you also forgot to put the percent sign "%"
    analyze_text = "The text contains {0} alphabetic characters, of which {1} ({2}%) are 'e'."

    # you don't need to add 1 to count and letter_count using += 1
    # they are already the correct values
    # also, you should return the string here, not print it
    return analyze_text.format(count, letter_count, p)

该代码可以为您提供您在问题中显示的所需结果

Answer 2

def analyze_text(text, letter='e'):
    n = len([x for x in text if x.isalpha()])
    freq = text.lower().count(letter)
    percent = float(freq) / n * 100
    return "The text contains {} alphabetic characters, of which {} ({}%) are '{}'.".format(n, freq, percent, letter)

Answer 3

快速浏览一下您的工作，看起来好像您错误地使用了替换字段。你的代码：

print(analyze_text.format(count += 1, letter_count += 1, p))

正确版本：

print(output.format(count, letter_count, p))

以下是您的功能应该是什么样的：

def analyze_text(text):
    count = 0
    letter_count = 0
    for char in text:
        if char.isalpha():
            count += 1
    for e in text:
        if e == "e" or e == "E":
            letter_count += 1
        p = float(letter_count)/float(count) * 100
    output = "The text contains {0} alphabetic characters, of which {1} ({2}%) are 'e'."
    print(output.format(count, letter_count, p))

您应该将格式化程序视为列表。您的替换字段对应于列表中的顺序。希望澄清一切。

P.S。 - 您忘记了输出消息中的％符号。

计数，找到某些字母和百分比

3 个答案: