字符串必须包含两个字符，而不仅仅是其中的一个

Question

可能听起来对你们来说太简单了，所以我很抱歉浪费你们的时间，但它让我可怜的大脑崩溃了。因此，就其价值而言，它是这样的：

第一个任务没问题：如果字符串有一个“i”或一个“u”，打印“有一个 i 或一个 u”。

string = "demo loops for you"
for char in string:
   if char == 'i' or char == 'u':
       print("There is an i or a u")

第二个是令人心碎的，尤其是在第一个条件下：

如果字符串同时具有“i”和“u”，则打印“同时存在 i 和 u”，因此如果只有“i”但没有“u”，则打印“只有 i”如果只有“u”但没有“i”，则打印“只有 u”

第一个条件的此块不起作用。它仍然打印（即使只有一个字符 - >“i”）：

string = "i can demo loops"
for char in string:
   if char == 'i' and 'u':
       print("There is both an i or u")

开导我哦，聪明人！ :'(

Answer 1

您可以使用两个计数变量来完成这项工作

string = "demo loops  u for yo"

icount=string.count('i')
ucount=string.count('u')


if (icount >0 and ucount==0) :
  print ("there is only i but not u")

if (icount ==0 and ucount>0):
  print ("there is only u but not i")

if (icount >0 and ucount>0):
  print ("there are both u and i")

Answer 2

首先，您可以检查字符串中的“i”和“u”，然后您可以一个一个地检查。

string = "i can demo loops u"

if "i" in string and "u" in string:
    print("There are both an i or u")
elif "i" in string :
    print("There is an i in string")
elif "u" in string :
    print("There is an u in string")

Answer 3

由于您提供的见解，我能够制作一个“稍微好一点”的版本，它不仅可以确定存在性，还可以要求输入、存储变量、计算字符数并相应地重点打印出值关于 singular/plural 名词。如果您看到我可能遗漏的内容，或者对如何使这段代码更“pythonic”有任何建议，我也将非常感激。再次感谢！

string = input ('enter a phrase, and i\'ll tell you how many i\'s and u\'s it has: ')
icounter = string.count('i')
ucounter = string.count('u')

if icounter == 0 and ucounter == 0:
    print ('you phrase has no i\'s or u\'s')
if icounter == 1 and ucounter == 1:
    print ('your phrase has', icounter , "i", "and", ucounter, "u")
if icounter > 1 and ucounter > 1:
    print ('your phrase has', icounter, "i's and", ucounter, "u's")
if icounter == 1 and ucounter == 0:
    print ('your phrase has an "i" but no "u"')
if icounter == 0 and ucounter == 1:
    print ('your phrase has a "u" but no "i"s')
if icounter > 1 and ucounter == 0:
    print ('your phrase has', icounter, '"i"\'s but no "u"\'s')
if icounter == 0 and ucounter > 1:
    print ('your phrase has', ucounter, '"u"\'s but no "i"\'s')
if icounter > 1 and ucounter == 1:
    print ('your phrase has', icounter, "i's and", 'a "u"')
if icounter == 1 and ucounter > 1:
    print ('your phrase has an "i"', "and", ucounter, 'u\'s')

字符串必须包含两个字符，而不仅仅是其中的一个

string MUST contain both chars and not just 1 of them

python

string

boolean

operators