给定一个字符串words和分隔符sep，请编写一个函数，统计字符串words中每个元素出现的次数，使用sep进行拼接。
示例：
输入：“hello world”，输出：" 1-1-3-3-2-1-1-2-1-3-1"题目难度：简单
题目来源：codewars: Frequency sequence
def freq_seq(words: str, sep: str) -> str:
pass
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
def freq_seq(words: str, sep: str) -> str:
times_dict = {}
for w in words:
if w not in times_dict.keys():
times_dict[w] = 1
else:
times_dict[w] += 1
return sep.join([str(times_dict[w]) for w in words])
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
def freq_seq(words:str,sep:str)->str:
temp_list=[]
for i in words:
count=words.count(i)
temp_list.append(str(count))
return sep.join(temp_list)
assert freq_seq("hello world", "-")=="1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") =="1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
def freq_seq(words: str, sep: str) -> str:
return sep.join([str(list(words).count(item)) for item in list(words) ])
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
整复杂啦
做题可以，业务代码这样写可读性就降低了。
def freq_seq(words: str, sep: str) -> str:
counts = {}
for i in words:
if i in counts.keys():
counts[i] += 1
else:
counts[i] = 1
return sep.join([str(counts[i]) for i in words])
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
def freq_seq(words: str, sep: str) -> str:
# 定义统计字典
counts = {}
# 将传入字符串转为列表
result = list(words)
# 遍历每个字符
for s in words:
# 统计每个字符次数
counts[s] = counts.get(s,0) + 1
# 遍历列表
for i in range(len(words)):
# 将每个字符替换成该字符统计的次数
result[i] = str(counts[result[i]])
# 通过分隔符将列表转为字符串
return sep.join(result)
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
def freq_seq(words: str, sep: str) -> str:
return sep.join([str(words.count(x)) for x in words])
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
思路：利用列表推导式，遍历字符串words，同时使用字符串count方法统计字符出现的次数，最后用join方法拼接整个列表的字符。
知识点：
count()，join()方法。def freq_seq(words: str, sep: str) -> str:
list1 = []
count_list = []
for i in words:
list1.append(i)
for j in list1:
count = str(list1.count(j))
if count is None:
pass
else:
count_list.append(count)
return f'{sep}'.join(count_list)
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"
public String freqSeq(String words,String step){
String[] split = words.split("");
String res ="";
LinkedHashMap<String,Integer> hs = new LinkedHashMap<>();
for (String s: split) {
if (hs.get(s) !=null){
hs.put(s, hs.get(s)+1);
}else {
hs.put(s, 1);
}
}
for (String s: hs.keySet()) {
res += hs.get(s)+step;
}
return res.substring(0, res.length()-1);
}
def freq_seq(words:str,sep:str):
res=[]
for i in words:
res.append(str(words.count(i)))
return sep.join(res)
def freq_seq(words: str, sep: str) -> str:
data = []
for i in words:
data.append([i, str(words.count(i))])
return f'{sep}'.join([x[1] for x in data])
assert freq_seq("hello world", "-") == "1-1-3-3-2-1-1-2-1-3-1"
assert freq_seq("19999999", ":") == "1:7:7:7:7:7:7:7"
assert freq_seq("^^^**$", "x") == "3x3x3x2x2x1"