如何查找列表中特定项目的索引？

如何解决如何查找列表中特定项目的索引？

我想在输出中打印带有索引号的项目，但我做不到。这是我的脚本的工作方式：

1.从用户那里获取一个字符串并删除句号和逗号。 2. 将字符串拆分为名为 reshte. 的列表。 3. 从索引 0 处的列表中删除第一项。 4. 扫描第一个章程为capital 的列表，然后在输出中显示。

输入：

波斯联赛是最大的体育赛事，致力于为伊朗的贫困地区。波斯联盟促进和平与友谊。这段视频是由我们的一位英雄拍摄的，他希望和平。

我的代码：

bs = [',','.']
reshte = input()
for s in bs:
    if s in reshte:
        reshte = reshte.replace(s,'')

reshte = reshte.split()
b = reshte[0]
css = [reshte.remove(b) for k in reshte if b in reshte]


for f in reshte:
    if f[0] == f[0].upper():
       print(f)

输出：

Persian
League
Iran
Persian
League

预期输出：

2:Persian
3:League
15:Iran
17:Persian
18:League

解决方法

@Patrick 确实给了你一个答案，但你说这是错误的，我假设索引的结果应该是索引 before 删除列表中的第一个元素（步骤 3）并将 1 添加到索引中，因为结果似乎是我们通常计算的索引，而不是 Python（或大多数编程语言，请参阅下面的指南）中的索引。

我还看到结果中没有“This”这个词，所以我假设您希望 ALSO 跳过所有以大写开头且与第一个元素相同的词列表，所以单词“The”。

1.使用列表作为结果

test_string = """The Persian League is the largest sport event dedicated,\
to the deprived,areas of Iran. The Persian League promotes,peace and friendship.  
This video was captured by one of our heroes,who wishes peace.
"""

bs = [',','.']
result_2 = []
# ========== < REMOVE COMMAS AND DOTS > ========= #
for punct in bs:
    if punct in test_string:
        test_string = test_string.replace(punct,'')


# ========== < STORE FIRST ELEMENT AT INDEX 0 > ========= #
reshte = test_string.split()
first_element = reshte[0]


# ========== < COUNTS ELEMENTS IN LIST > ========= #
results = []
for idx,f in enumerate(reshte):
    if first_element[0] == f[0]: # NOTE: we don't remove the first element but only skipping it all in once
        continue
    if f[0].isupper():
        results.append((idx,f))

for idx,value in results:
    print(f'{idx+1}:{value}')

# 2:Persian
# 3:League
# 15:Iran
# 17:Persian
# 18:League

2.使用字典

results = dict()
for idx,f in enumerate(reshte):
    if first_element[0] == f[0]: # NOTE: we don't remove the first element
        continue
    if f[0].isupper():
        results[idx] = f

for key,value in results.items():
    print(f'{key+1}:{value}')
    
# 2:Persian
# 3:League
# 15:Iran
# 17:Persian
# 18:League

print(results)

# {1: 'Persian',2: 'League',14: 'Iran',16: 'Persian',17: 'League'}

3.计数重复的列表

# ========== < FIND UNIQUE ELEMENT IN LIST > ========= #
unique_string = {string for string in reshte if string[0].isupper()}
results = []
# ========== < COUNT ELEMENTS  > ========= #
for element in unique_string:
    if first_element[0] == element[0]:
        continue
    element_count = reshte.count(element)
    results.append((element_count,element))

for idx,value in results:
    print(f'{idx}:{value}')
# 1: Iran
# 2: League
# 2: Persian

4.使用计数器

from collections import Counter

counted = Counter(reshte)

for element,count in counted.items():
    if element[0] == first_element[0]:
        continue
    elif element[0].isupper():
        print(f'{count}:{element}')
        
# 2:Persian
# 2:League
# 1:Iran

指南和文档

堆栈溢出相关问题

使用 enumerate 和 print 的 sep 选项解决这个问题。只需更改最后一个 for 循环。

for i,f in enumerate(reshte):
    if f[0] == f[0].upper():
       print(i,f,sep=":")