計數，發現某些字母和百分比

Question

我正在處理這個問題，但找不到正確的答案，但我設法挖了一個更大的洞&迷惑自己。因此，如果任何人都可以提供清晰：

方向：

寫接收一個字符串作爲輸入的功能analyze_text。您的函數應計算文本中的字母字符數（a到z或從A到Z），並記錄字母'e'（大寫或小寫）的數量。

你的功能應該在一個字符串的確切措辭這樣的形式返回文本的分析：「文本包含240個字母字符，其中105（43.75％）的是‘E’」

您將需要使用isalpha函數。

到目前爲止我的代碼： 高清analyze_text（文本）： 計數= 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

有錯誤的評論概述如下只有兩件事情：

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) 

該代碼應該爲您提供您在問題中顯示的所需結果

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

快速瀏覽一下您的工作，看起來好像您錯誤地使用了替換字段。您的代碼：

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

正確的版本：

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

這是你的功能應該是什麼樣子：

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)) 

你應該把你的格式化爲列表。您的替換字段與列表中的訂單相對應。希望澄清事情。

P.S. - 您在輸出訊息中忘記了％符號。