我正在计算原始输入'短语'中有多少o并计算第一个o并表示已经完成但是之后有很多o。我如何通过no for循环来计算所有的o。
while loop_count<len(phrase):
if "o" in phrase:
loop_count += 1
count2 += 1
if loop_count>len(phrase):
print loop_count
break
else:
continue
else:
print loop_count
continue
答案 0 :(得分:2)
您可以将sum与迭代器一起使用(在本例中为generator expression):
>>> sum(c=='o' for c in 'oompa loompa')
4
你可以使用带有len的正则表达式:
>>> re.findall('o', 'oompa loompa')
['o', 'o', 'o', 'o']
>>> len(re.findall('o', 'oompa loompa'))
4
您可以使用计数器:
>>> from collections import Counter
>>> Counter('oompa loompa')['o']
4
或者只使用字符串的'count'方法:
>>> 'oompa loompa'.count('o')
4
如果您确实希望使用while循环,请使用pop方法将列表作为堆栈使用:
s='oompa loompa'
tgt=list(s)
count=0
while tgt:
if tgt.pop()=='o':
count+=1
或'for'循环 - 更多Pythonic:
count=0
for c in s:
if c=='o':
count+=1
答案 1 :(得分:2)
您可以使用count
功能:
phrase.count('o')
但是如果你想在'o'的一个匹配之后发送一条消息,对于所有字符串的跳过循环,只需使用'in',如下所示:
if 'o' in phrase :
# show your message ex: print('this is false')
答案 2 :(得分:1)
让我们尝试剖析您了解正在发生的事情的代码。看我的评论
while loop_count<len(phrase):
if "o" in phrase: # See comment 1
loop_count += 1 # Why are you incrementing both of these?
count2 += 1
if loop_count>len(phrase): # What is this statement for? It should never happen.
## A loop always exits once the condition at the while line is false
## It seems like you are manually trying to control your loop
## Which you dont need to do
print loop_count
break # What does this accomplish?
else:
continue
else:
print loop_count
continue # Pointless as we are already at end of loop. Consider removing
评论1:您在询问是否有&#39; o&#39;在这个短语中的任何地方。你想要询问当前的字母是否是o。也许您打算使用索引访问短语的字母,例如if 'o' == phrase[loop_count]
。如果你这样做,你想每次都增加loop_count,但只有在字母是o时才算数。
您可以像这样重写:
loop_count, o_count = 0, 0
while loop_count<len(phrase):
# loop_count represents how many times we have looped so far
if "o" == phrase[loop_count].lower(): # We ask if the current letter is 'o'
o_count += 1 # If it is an 'o', increment count
loop_count += 1 # Increment this every time, it tracks our loop count
print o_count
答案 3 :(得分:0)
使用理解
len([i for i in phrase if i=="o"])