計算字母數字的Python函數，記錄多少次「e」出現

Question

編寫一個函數analyze_text，它接收一個字符串作爲輸入。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。

該函數應返回文本的分析，是這樣的：

文本包含240個字母字符，其中105（43.75％）的是「E」。

我需要使用isalpha功能，它可以這樣使用的：

"a".isalpha() # => evaluates to True 
"3".isalpha() # => evaluates to False 
"&".isalpha() # => False 
" ".isalpha() # => False 

mystr = "Q" 
mystr.isalpha() # => True 

功能應通過以下測試：

from test import testEqual 

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) 

所以，我想：

def analyze_text(text): 

    text = input("Enter some text") 
    alphaChars = len(text) 

#count the number of times "e" appears 
    eChars = text.count('e') 

#find percentage that "e" appears 
    eCharsPercent = eChars/alphaChars 

    print("The text contains" + alphaChars + "alphabetic characters, of 
    which" + eChars + "(" + eCharsPercent + ") are 'e'.") 

from test import testEqual 

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) 

正如你所看到的，我嘗試過的並不使用isalpha函數（我不知道如何/在哪裏使用我T）。此外，無論測試是否通過，函數都不會返回。可視化python不支持「測試」，我在書中使用的文本編輯器說我有縮進錯誤（？）我不知道從哪裏開始 - 請幫助。

Screenshot of Book Text Editor

EDIT：現在接收 「類型錯誤：不能連接於線12 'STR' 和 'INT' 對象」（即與 「打印」 開頭的行）。

Answer 1

這裏是一個analyse_text功能通過測試：

def analyze_text(text): 
    filtered = [c.lower() for c in text if c.isalpha()] 
    cnt = filtered.count('e') 
    result = "The text contains {} alphabetic characters, of which {} ({}%) are 'e'.".format(len(filtered),cnt,str(100.0*cnt/len(filtered))[:13]) 
    return result 

• 創建所有alphanum字符，小寫的列表，用一個漂亮的列表理解（也就是你失蹤的部分），創建filtered變量
• count count e（you got that right），創建cnt計數器
• 格式字符串accordignly（使用了一點破解得到33.3333333 ri ght，或許可以做更好的事情）。創建result字符串後返回