[‘AC’, ‘2H’, ‘3S’, ‘4C’]
如何检查每个字符串的第一个索引(例如第二个元素)是否多次出现?例如,在这种情况下,C发生2次,因此我需要返回False 这也必须适用于其他情况,例如H或S多次出现
答案 0 :(得分:1)
请考虑使用collections.Counter
对感兴趣的项目进行计数。然后使用all
或any
验证条件。
import collections
a = ['AC', '2H', '3S', '4C']
counter = collections.Counter(s[1] for s in a)
result = all(v < 2 for v in counter.values())
print(result)
答案 1 :(得分:0)
您可以使用此功能:
def check_amount(all_text, character):
count = 0
for text in all_text:
for ch in text:
if ch == character:
count += 1
return count
如果您只想查看它是否存在,则返回它发生的次数:
def check_amount(all_text, character):
for text in all_text:
for ch in text:
if ch == character:
return True
else:
return False
这些都是用于检查任何位置的,如果您需要将其放在特定位置,如您所说:
def check_amount(all_text, character):
count = 0
for text in all_text:
if text[1] == character:
count += 1
return count
然后,如果要使用不使用计数的相同方法来获取布尔版本,则可以更改此值
all_text
是您要传递的列表,而character
是您要查看的列表/是否存在。
答案 2 :(得分:0)
使用正则表达式,您可以使用re.finditer
查找所有(非重叠)事件:
>>> import re
>>> text = 'Allowed Hello Hollow'
>>> for m in re.finditer('ll', text):
print('ll found', m.start(), m.end())
ll found 1 3
ll found 10 12
ll found 16 18
或者,如果您不想正则表达式的开销,也可以重复使用str.find来获取下一个索引:
>>> text = 'Allowed Hello Hollow'
>>> index = 0
>>> while index < len(text):
index = text.find('ll', index)
if index == -1:
break
print('ll found at', index)
index += 2 # +2 because len('ll') == 2
ll found at 1
ll found at 10
ll found at 16
This also works for lists and other sequences.
对于这里的数组,我将使用列表推导,如下所示:
listOfElems = ['Hello', 'Ok', 'is', 'Ok', 'test', 'this', 'is', 'a', 'test', 'Ok']
现在让我们在列表中找到所有'ok'的索引
# Use List Comprehension Get indexes of all occurrences of 'Ok' in the list
indexPosList = [ i for i in range(len(listOfElems)) if listOfElems[i] == 'Ok' ]
print('Indexes of all occurrences of "Ok" in the list are: ', indexPosList)
输出:
Indexes of all occurrences of "Ok" in the list are : [1, 3, 9]