问题描述
程序的工作原理
3.split()
将该字符串放入名为 reshte
的列表中。
我想在输出中打印带有索引号的项目,但我不能。
我给程序的输入:
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)
Persian
League
Iran
Persian
League
但我希望输出如下==正确:
2:Persian
3:League
15:Iran
17:Persian
18:League
请帮我解决这个问题
解决方法
只需更改最后一部分即可打印数字:
const DEFAULT_NUMBER_VALUE = 0;
const DEFAULT_PRECISION = 100;
function parseAddInput(input){
if (!input) {
return DEFAULT_NUMBER_VALUE;
}
if (typeof input === 'string'){
input = parseInt(input);
}
const roundedNumber = Math.round(input * (10 * DEFAULT_PRECISION));
return roundedNumber;
}
function getSum(a,b){
return Array(
parseAddInput(a)
).concat(
Array(parseAddInput(b))
).length / 100;
}
function add(number1,number2){
return getSum(number1,number2);
}
,
问题在于,当您从列表中删除项目时,您输出的单词的索引会发生变化。最简单的做法是之后删除项目,以便在循环中,单词仍保留其原始索引。这样做时,您可以简单地在循环中添加条件 f != b
以过滤掉 reshte[0]
不被打印。
以下代码实现了这一点。请注意,您的代码的第一部分可以大大简化 - 在进行替换之前无需检查 ,
或 .
是否确实出现在字符串中 - replace
是空操作如果文本没有出现在字符串中。
reshte = input()
reshte = reshte.replace(',','').replace('.','')
reshte = reshte.split()
b = reshte[0]
for i,f in enumerate(reshte):
if f[0] == f[0].upper() and f != b:
print(f'{i + 1}:{f}')
css = [reshte.remove(b) for k in reshte if b in reshte]
,
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)并向索引添加 1,因为结果看起来是我们的索引通常计算,而不是在 Python(或大多数编程语言,请参阅下面的指南)中计算。
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
- python_access_list_items
- geeksforgeeks.org | isupper-islower-lower-upper
- Python Docs | collections.Counter