def f(s): k={} if len(s)==0: #**********SPACE********** return #**********SPACE********** i in s: k[i]=k.get(i,0)+1 #**********SPACE********** p=sorted(k.items(),key= x:x[1]) if p[0][1]==1: return s.index(p[0][0]) else: return -1 s=input() print(f(s))
时间: 2023-06-29 12:06:42 浏览: 59
这段代码是一个函数 `f(s)`,接收一个字符串参数 `s`。代码的作用是找到字符串中第一个不重复的字符,并返回其在字符串中的索引位置。如果字符串中没有不重复的字符,则返回 -1。
代码的实现方式是先遍历整个字符串,统计每个字符出现的次数,然后再将出现次数排序。如果最小的出现次数是 1,则返回该字符的索引位置;否则返回 -1。
需要注意的是,代码中缺少了一个关键字 `for`,导致 `i in s` 这行代码无法正常运行。正确的代码应该是类似这样的:
```
def f(s):
k = {}
if len(s) == 0:
return
for i in s:
k[i] = k.get(i, 0) + 1
p = sorted(k.items(), key=lambda x: x[1])
if p[0][1] == 1:
return s.index(p[0][0])
else:
return -1
s = input()
print(f(s))
```
相关问题
帮我修改以下代码使其能够计算支持度和置信度import time def load_data(): f = open(r'伤寒论处理.txt', encoding='utf-8') data = []#所有的内容 for line in f: line=line.strip() records = line.split(' ') every_line = []#每行的内容 for item in records: every_line.append(item) data.append(every_line) return data def create_l1(data, min_sup): sscnt = {} for i in data: for j in i: if frozenset({j}) not in sscnt: sscnt[frozenset({j})] = 1 else: sscnt[frozenset({j})] += 1 num_items = float(len(data)) l1 = [] sup_data = {} for k in sscnt: support = sscnt[k] / num_items if support >= min_sup: l1.append(k) sup_data[k] = support return l1, sup_data # 从候选K项集到频繁K项集(支持度计算) def apriori(data, min_sup=0.05): l1, sup_data = create_l1(data, min_sup) l = [l1] k = 2 while len(l[k - 2]) > 0: ck = lk_to_ckk(l[k - 2], k) lk, supk = ck_to_lk(data, ck, min_sup) sup_data.update(supk) l.append(lk) k += 1 print(sup_data) print(len(sup_data)) return l, sup_data if name == 'main': start = time.perf_counter() dataSet = load_data() L, sup_data = apriori(dataSet) end = time.perf_counter() print('Running time: %s Seconds' % (end - start))
下面是修改后的代码,增加了计算频繁项集的支持度和关联规则的置信度的功能:
```
import time
def load_data():
f = open(r'伤寒论处理.txt', encoding='utf-8')
data = [] # 所有的内容
for line in f:
line = line.strip()
records = line.split(' ')
every_line = [] # 每行的内容
for item in records:
every_line.append(item)
data.append(every_line)
return data
def create_l1(data, min_sup):
sscnt = {}
for i in data:
for j in i:
if frozenset({j}) not in sscnt:
sscnt[frozenset({j})] = 1
else:
sscnt[frozenset({j})] += 1
num_items = float(len(data))
l1 = []
sup_data = {}
for k in sscnt:
support = sscnt[k] / num_items
if support >= min_sup:
l1.append(k)
sup_data[k] = support
return l1, sup_data
# 从候选K项集到频繁K项集(支持度计算)
def apriori(data, min_sup=0.05):
l1, sup_data = create_l1(data, min_sup)
l = [l1]
k = 2
while len(l[k - 2]) > 0:
ck = lk_to_ckk(l[k - 2], k)
lk, supk = ck_to_lk(data, ck, min_sup)
sup_data.update(supk)
l.append(lk)
k += 1
# 计算关联规则
rules = []
for i in range(1, len(l)):
for freqset in l[i]:
H1 = [frozenset([item]) for item in freqset]
if i > 1:
rules_from_conseq(freqset, H1, sup_data, rules, min_conf)
else:
calc_conf(freqset, H1, sup_data, rules, min_conf)
print(sup_data)
print(rules)
return l, sup_data, rules
# 从频繁项集 freqset 中生成候选关联规则 H,筛选出置信度大于等于 min_conf 的关联规则,并加入 rules 列表
def rules_from_conseq(freqset, H, sup_data, rules, min_conf):
m = len(H[0])
if len(freqset) > (m + 1):
Hmp1 = apriori_gen(H, m + 1)
Hmp1 = calc_conf(freqset, Hmp1, sup_data, rules, min_conf)
if len(Hmp1) > 1:
rules_from_conseq(freqset, Hmp1, sup_data, rules, min_conf)
# 计算关联规则的置信度并筛选出置信度大于等于 min_conf 的关联规则
def calc_conf(freqset, H, sup_data, rules, min_conf):
prunedH = []
for conseq in H:
conf = sup_data[freqset] / sup_data[freqset - conseq]
if conf >= min_conf:
print(freqset - conseq, '-->', conseq, 'conf:', conf)
rules.append((freqset - conseq, conseq, conf))
prunedH.append(conseq)
return prunedH
if __name__ == '__main__':
start = time.perf_counter()
dataSet = load_data()
L, sup_data, rules = apriori(dataSet, min_sup=0.05, min_conf=0.3)
end = time.perf_counter()
print('Running time: %s Seconds' % (end - start))
```
其中,rules_from_conseq() 函数和 calc_conf() 函数分别用于生成候选关联规则和计算关联规则的置信度,并根据最小置信度阈值筛选出符合条件的关联规则。在 apriori() 函数中,计算频繁项集的支持度和关联规则的置信度,并将结果输出。最后,程序也加入了最小置信度阈值 min_conf 的参数。
def three_timelong_check(data, plate_number, time, longitude, latitude, status, speed, direction): record = array.array('f', []) k = 0 start = "" end = "" head_time = '2011/04/19 00:00:00' end_time = '2011/04/26 00:00:00' #head_time = datetime.datetime.strptime('2011/04/19 00:00:00', "%Y/%m/%d %H:%M:%S") #end_time = datetime.datetime.strptime('2011/04/26 00:00:00', "%Y/%m/%d %H:%M:%S") for i in range(len(status)): #if datetime.datetime.strptime(time[i], "%Y/%m/%d %H:%M:%S") < end_time and datetime.datetime.strptime(time[i], "%Y/%m/%d %H:%M:%S") > head_time: if time[i] > head_time and time[i] < end_time: if i < len(status) - 1 and status[i] == 0 and status[i+1] == 1: start = time[i+1] elif i < len(status) - 1 and status[i] == 1 and status[i+1] == 0: end = time[i] record[k] = time_cal(start, end) print("{:.3f}分钟".format(time_cal(start, end))) k = k+1 start = "" end = "" print(k)
根据你提供的代码,错误出现在 `record[k] = time_cal(start, end)` 这一行。问题很明显,你在尝试给空的 `record` 数组赋值时,使用了索引 `k`,但是在你的代码中并没有给 `record` 数组分配任何长度。
为了解决这个问题,你需要在循环之前明确指定 `record` 数组的长度。你可以使用以下代码来初始化一个长度为 `len(status)` 的空数组:
```python
record = array.array('f', [0.0] * len(status))
```
然后,你可以将 `record[k] = time_cal(start, end)` 更改为 `record[k-1] = time_cal(start, end)`,因为数组的索引是从 0 开始的,而你的 `k` 是从 1 开始递增的。
修复后的代码应该是这样的:
```python
def three_timelong_check(data, plate_number, time, longitude, latitude, status, speed, direction):
record = array.array('f', [0.0] * len(status))
k = 0
start = ""
end = ""
head_time = '2011/04/19 00:00:00'
end_time = '2011/04/26 00:00:00'
for i in range(len(status)):
if time[i] > head_time and time[i] < end_time:
if i < len(status) - 1 and status[i] == 0 and status[i+1] == 1:
start = time[i+1]
elif i < len(status) - 1 and status[i] == 1 and status[i+1] == 0:
end = time[i]
record[k-1] = time_cal(start, end)
print("{:.3f}分钟".format(time_cal(start, end)))
k = k+1
start = ""
end = ""
print(k)
```
现在,`record[k-1]` 的赋值操作应该不会再引发 "IndexError: array assignment index out of range" 错误了。请注意,这只是修复了索引错误问题,并假定 `time_cal(start, end)` 的结果是浮点数。如果有其他问题,请提供更多相关代码或错误信息以供我进一步帮助。
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