计数，找到某些字母和百分比答案

【问题标题】：Counting, finding certain letter and percentage计数，找到某些字母和百分比
【发布时间】：2017-08-14 23:19:35
【问题描述】：

我正在研究这个问题，找不到正确的答案，但我设法挖了一个更大的洞并让自己感到困惑。因此，如果有人可以提供清晰的说明：

方向：

编写一个函数analyze_text，它接收一个字符串作为输入。您的函数应计算文本中字母字符（a 到 z 或 A 到 Z）的数量，并跟踪字母“e”（大写或小写）的数量。

您的函数应该以字符串的形式返回对文本的分析，其措辞完全如下：

“文本包含 240 个字母字符，其中 105 个（43.75%）是‘e’。”

您需要使用 isalpha 函数。

到目前为止我的代码： def 分析文本（文本）：计数 = 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)

【问题讨论】：

你需要展示你遇到了什么错误，否则你不会得到任何帮助，但你会得到很多反对票
非常接近，但您不需要打印答案字符串，而是需要返回它。此外，通过在同一个循环中增加 count 和 letter_count 可以提高效率。

标签： python string percentage alpha

【解决方案1】：

下面的 cmets 中只有几处错误：

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)

该代码应该为您提供问题中显示的所需结果

【讨论】：

【解决方案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)

【讨论】：

【解决方案3】：

快速浏览一下您的工作，您似乎在错误地使用替换字段。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))

你的函数应该是这样的：

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一个列表。您的替换字段对应于列表中的顺序。希望澄清一下。

附： - 您忘记了输出消息中的 % 符号。

【讨论】：