我遇到一个问题，明确要求我不要使用numpy或pandas。
问题：
给出一个包含数字和'_'（缺失值）符号的字符串，您必须按照说明替换'_'符号
Ex 1: _, _, _, 24 ==> 24/4, 24/4, 24/4, 24/4 i.e we. have distributed the 24 equally to all 4 places
Ex 2: 40, _, _, _, 60 ==> (60+40)/5,(60+40)/5,(60+40)/5,(60+40)/5,(60+40)/5 ==> 20, 20, 20, 20, 20 i.e. the sum of (60+40) is distributed qually to all 5 places
Ex 3: 80, _, _, _, _ ==> 80/5,80/5,80/5,80/5,80/5 ==> 16, 16, 16, 16, 16 i.e. the 80 is distributed qually to all 5 missing values that are right to it
Ex 4: _, _, 30, _, _, _, 50, _, _
==> we will fill the missing values from left to right
a. first we will distribute the 30 to left two missing values (10, 10, 10, _, _, _, 50, _, _)
b. now distribute the sum (10+50) missing values in between (10, 10, 12, 12, 12, 12, 12, _, _)
c. now we will distribute 12 to right side missing values (10, 10, 12, 12, 12, 12, 4, 4, 4)
将同时具有两个缺失值，例如：“ _，_，x，_，_，”，您需要填写缺失值Q：您的程序读取字符串例如ex：“ ，_，x，_，_，_”，并返回填充的序列Ex：
Input1: "_,_,_,24"
Output1: 6,6,6,6
Input2: "40,_,_,_,60"
Output2: 20,20,20,20,20
Input3: "80,_,_,_,_"
Output3: 16,16,16,16,16
Input4: "_,_,30,_,_,_,50,_,_"
Output4: 10,10,12,12,12,12,4,4,4
我正在尝试使用split函数将字符串拆分为列表。然后，我尝试检查左侧的空白，并计算此类空白的数量，然后一旦遇到非空白，就用该数字除以总计数，即（在数字之前遇到的空白和数字本身），然后散布值并替换空格，剩下数字
然后我要检查两个数字之间的空格，然后应用相同的逻辑，然后对右侧的空格进行相同的操作。
但是，我在下面共享的代码引发了各种各样的错误，并且我相信我在上面共享的逻辑上存在空白，因此希望能对解决此问题有深刻的见识
def blanks(S):
a= S.split()
count = 0
middle_store = 0
#left
for i in range(len(a)):
if(a[i]=='_'):
count = count+1 #find number of blanks to the left of a number
else:
for j in range(0,i+1):
#if there are n blanks to the left of the number speard the number equal over n+1 spaces
a[j] = str((int(a[i])/(count+1)))
middle_store= i
break
#blanks in the middle
denominator =0
flag = 0
for k in len(middle_store+1,len(a)):
if(a[k] !='_'):
denominator = (k+1-middle_store)
flag=k
break
for p in len(middle_store,flag+1):
a[p] = str((int(a[p])/denominator))
#blanks at the right
for q in len(flag,len(a)):
a[q] = str((int(a[q])/(len(a)-flag+1)))
S= "_,_,30,_,_,_,50,_,_"
print(blanks(S))
答案 0 :(得分：2)
首先，您应该在split方法中指定一个定界符作为参数，默认情况下，它将按空格分隔。
所以
"_,_,x,_,_,y,_".split()给您['_,_,x,_,_,y,_']
而"_,_,x,_,_,y,_".split(',')会给您['_', '_', 'x', '_', '_', 'y', '_']。
第二，对于“中间”和“右”循环（对于右），您需要将len替换为range。
由于要进行除法，因此最好使用float而不是int
由于将其用于除法，因此最好将分母初始化为1。
在最后一个循环中，a[q] = str((int(a[q])/(len(a)-flag+1)))（与a[p]相同）应返回错误，因为a [q]为“ _”。您需要使用变量来保存a[flag]值。
每个中断都应在else或if语句中，否则，您将只通过一次循环。
最后，为了提高复杂度，您可以从j循环中退出middle_store分配，以避免每次都对其进行分配。
TL; DR：试试这个：
def blanks(S):
a = S.split(',')
count = 0
middle_store = 0
# left
for i in range(len(a)):
if a[i] == '_':
count = count + 1 # find number of blanks to the left of a number
else:
for j in range(i + 1):
# if there are n blanks to the left of the number speard the number equal over n+1 spaces
a[j] = str((float(a[i]) / (count + 1)))
middle_store = i
middle_store_value = float(a[i])
break
# blanks in the middle
denominator = 1
flag = 0
for k in range(middle_store + 1, len(a)):
if a[k] != '_':
denominator = (k + 1 - middle_store)
flag = k
break
flag_value = float(a[flag])
for p in range(middle_store, flag + 1):
a[p] = str((middle_store_value+flag_value) / denominator)
# blanks at the right
last_value = float(a[flag])
for q in range(flag, len(a)):
a[q] = str(last_value / (len(a) - flag))
return a
S= "_,_,30,_,_,_,50,_,_"
print(blanks(S))
PS：您甚至尝试过解决错误吗？还是只是等待某人解决您的数学问题？
答案 1 :(得分：1)
模块化解决方案
# takes an array x and two indices a,b.
# Replaces all the _'s with (x[a]+x[b])/(b-a+1)
def fun(x, a, b):
if a == -1:
v = float(x[b])/(b+1)
for i in range(a+1,b+1):
x[i] = v
elif b == -1:
v = float(x[a])/(len(x)-a)
for i in range(a, len(x)):
x[i] = v
else:
v = (float(x[a])+float(x[b]))/(b-a+1)
for i in range(a,b+1):
x[i] = v
return x
def replace(text):
# Create array from the string
x = text.replace(" ","").split(",")
# Get all the pairs of indices having number
y = [i for i, v in enumerate(x) if v != '_']
# Starting with _ ?
if y[0] != 0:
y = [-1] + y
# Ending with _ ?
if y[-1] != len(x)-1:
y = y + [-1]
# run over all the pairs
for (a, b) in zip(y[:-1], y[1:]):
fun(x,a,b)
return x
# Test cases
tests = [
"_,_,_,24",
"40,_,_,_,60",
"80,_,_,_,_",
"_,_,30,_,_,_,50,_,_"]
for i in tests:
print (replace(i))
答案 2 :(得分：0)
针对所引发问题的代码也可以通过以下方式来完成，尽管该代码并未经过优化和简化，但它是从不同的角度编写的：
import re
def curve_smoothing(S):
pattern = '\d+'
ls_num=re.findall(pattern, S) # list of numeral present in string
pattern = '\d+'
spaces = re.split(pattern, S) # split string to seperate '_' spaces
if len(spaces[0])==0 and len(ls_num)==1:
Space_num=len(re.findall('_', S))
sums=int(ls_num[0])
repl_int=round(sums/(Space_num+1))
S=re.sub(r'(\d{2})(,_)', r'{}\2'.format(str(repl_int)) , S, 1)
S=re.sub('_', str(repl_int),S, Space_num)
return S
elif len(spaces[0])==0 and len(ls_num)>1:
for i in range(1,len(spaces)):
if i==1:
Space_num=len(re.findall('_', spaces[i]))
sums=int(ls_num[i-1])+int(ls_num[(i)])
repl_int=round(sums/(Space_num+2))
S=re.sub(str(ls_num[i-1]), str(repl_int),S)
S=re.sub('_', str(repl_int),S, Space_num)
S=re.sub(str(ls_num[i]), str(repl_int),S,1)
ls_num[i]=repl_int
elif i<len(ls_num):
Space_num=len(re.findall('_', spaces[i]))
sums=int(ls_num[i-1])+int(ls_num[(i)])
repl_int=round(sums/(Space_num+2))
S=re.sub(r'(\d{2})(,_)', r'{}\2'.format(str(repl_int)) , S, 1)
S=re.sub('_', str(repl_int),S, Space_num)
S=re.sub(str(ls_num[i]), str(repl_int),S,1)
ls_num[i]=repl_int
elif len(spaces[-1])!=0:
Space_num=len(re.findall('_', spaces[i]))
repl_int=round(ls_num[(i-1)]/(Space_num+1))
S=re.sub(r'(\d{2})(,_)', r'{}\2'.format(str(repl_int)) , S, 1)
S=re.sub('_', str(repl_int),S, Space_num)
return S
else:
for i in range(len(spaces)):
if i==0:
Space_num=len(re.findall('_', spaces[i]))
sums=int(ls_num[(i)])
repl_int=round(sums/(Space_num+1))
S=re.sub(r'(\d{2})(,_)', r'{}\2'.format(str(repl_int)) , S, 1)
S=re.sub('_', str(repl_int),S, Space_num)
S=re.sub(str(ls_num[i]), str(repl_int),S, 1)
ls_num[i]=repl_int
elif i>=1 and i<len(ls_num):
Space_num=len(re.findall('_', spaces[i]))
sums=int(ls_num[i-1])+int(ls_num[(i)])
repl_int=round(sums/(Space_num+2))
S=re.sub(r'(\d{2})(,_)', r'{}\2'.format(str(repl_int)) , S, 1)
S=re.sub('_', str(repl_int),S, Space_num)
S=re.sub(str(ls_num[i]), str(repl_int),S,1)
ls_num[i]=repl_int
elif len(spaces[-1])!=0:
Space_num=len(re.findall('_', spaces[i]))
repl_int=round(ls_num[(i-1)]/(Space_num+1))
S=re.sub(r'(\d{2})(,_)', r'{}\2'.format(str(repl_int)) , S, 1)
S=re.sub('_', str(repl_int),S, Space_num)
return S
S1="_,_,_,24"
S2="40,_,_,_,60"
S3= "80,_,_,_,_"
S4="_,_,30,_,_,_,50,_,_"
S5="10_,_,30,_,_,_,50,_,_"
S6="_,_,30,_,_,_,50,_,_20"
S7="10_,_,30,_,_,_,50,_,_20"
print(curve_smoothing(S1))
print(curve_smoothing(S2))
print(curve_smoothing(S3))
print(curve_smoothing(S4))
print(curve_smoothing(S5))
print(curve_smoothing(S6))
print(curve_smoothing(S7))
答案 3 :(得分：0)
# _, _, 30, _, _, _, 50, _, _
def replace(string):
lst=string.split(',')
for i in range(len(lst)):
if lst[i].isdigit():
for j in range(i+1):
lst[j]=int(lst[i])//(i+1)
new_index=i
new_value=int(lst[i])
break
for i in range(new_index+1,len(lst)):
if lst[i].isdigit():
temp=(new_value+int(lst[i]))//(i-new_index+1)
for j in range(new_index,i+1):
lst[j]=temp
new_index=i
new_value=int(lst[i])
try:
for i in range(new_index+1,len(lst)):
if not(lst[i].isdigit()):
count=lst.count('_')
break
temp1=new_value//(count+1)
for i in range(new_index,len(lst)):
lst[i]=temp1
except:
pass
return lst
答案 4 :(得分：0)
def replace(string):
lst=string.split(',')
if lst[0].strip().isdigit():
index0=0
while True:
index1=index0
value1=int(lst[index0].strip())
index2=index1
for i in range((index1+1),len(lst)):
if lst[i].strip().isdigit():
index2=i
break
value2=0
if index2>index1:
value2=int(lst[index2].strip())
else:
index2=len(lst)-1
value=str(int((value1+value2)/((index2-index1)+1)))
for i in range(index1,index2+1):
lst[i]=value
index0=index2
if index0>=(len(lst)-1):
break
else:
index0=0
while True:
index1=index0
value1=0
if lst[index0].strip().isdigit():
value1=int(lst[index0].strip())
index2=index1
for i in range((index1+1),len(lst)):
if lst[i].strip().isdigit():
index2=i
break
value2=0
if index2>index1:
value2=int(lst[index2].strip())
else:
index2=len(lst)-1
value=str(int((value1+value2)/((index2-index1)+1)))
for i in range(index1,index2+1):
lst[i]=value
index0=index2
if index0>=(len(lst)-1):
break
return lst
string = "20,_,_,30, _, _,10,_,_,_,_,110"
replace(string)
答案 5 :(得分：-1)
“检查所有输入的内容是否有效”
def替换：
val=0
lst=s.split(",")
if lst[0].isdigit():
for i in range(1,len(lst)):
if lst[i].isdigit():
value=(int(lst[0])+int(lst[i]))//((i+1))
for j in range(0,i+1):
lst[j]=value
index=i
break
else:
for i in range(len(s)):
if lst[i].isdigit():
for j in range(i+1):
lst[j]=(int(lst[i]))//(i+1)
index=i
value=int(lst[i])
break
for i in range(index+1,len(lst)):
if lst[i].isdigit():
temp=(value+int(lst[i]))//(i-index+1)
for j in range(index,i+1):
lst[j]=temp
index=i
value=int(lst[i])
try :
for i in range(index+1,len(lst)):
if not(lst[i].isdigit()):
count=lst.count('_')
break
temp1=value//(count+1)
for i in range(index,len(lst)):
lst[i]=temp1
except UnboundLocalError as e:
print (e)
return lst