简体   繁体   中英

select specific items from a list

I am having the following as my input:

a10=[['p', 'e'], ['t', 'h','a'],['e', 'a'],['p','e'],['a']]

c10=[['p','e'],['e','h'],['e', 'a']]

a11=[4,5,2,1,3]

average is calculated in this way:

first it should check if first element of c10 ie ['p','e'] is present in a10 or not. If its present, then it should take all the matching indexes(here indexes 0 and 3). Now it should go to those index positions in 'a11' and compute the mean, so it will be (4+1)/2=2.5

If there is no exact match, say for second element of c10 ie ['e','h'] , then it should check with individual element, here 'e' and 'h' and search for the indexes in which these charachters are present in a10 , here indexes are: 0,2,3 (since in these positions 'e' is present) and 1(since in these positions 'h' is present) and go for these indexes in a11 and compute average. so average=(4+5+2+1)/4

Hence, My output should be like this:

average=2.5 #ie average=(4+1)/2 --so if there is an exact match it should take the respective element from 'a11'..do the sum and compute mean.

average=3.0 #ie average=(4+5+2+1)/4 --if there is no exact match, it should match for the individual elements in the list of list and then compute mean by taking values in a11 .

average=2.0 #ie average=2 --if there is an exact match, it should return that value.

I have written the following code:

  a10=[['p', 'e'], ['t', 'h','e'],['e', 'a'],['p','e'],['a']]

  c10=[['p','e'],['e','h'],['e', 'a']]

  a11=[4,5,2,1,3]

  max_=3
  for i in range(len(a10)):
      if len(a10[i])<3:
          a10[i]=a10[i]+(3-len(a10[i]))*[str('')]

  for i in range(len(c10)):
      if len(c10[i])<3:
          c10[i]=c10[i]+(3-len(c10[i]))*[str('')]


  for w in range(len(c10)):
      total1=0
      count1=0
      for i in range(len(a10)):
          if c10[w] in a10:
              total1=total1+a11[i]
              count1=count1+1            
              average=float(total1/count1)
              #break

          else:
              total2=0
              count2=0
              for i in range(len(a10)):
                  for j in range(len(a10[i])):
                      for k in range(len(c10[w])):
                          if c10[w][k]==a10[i][j]:
                              total2=total2+a11[i]
                              count2=count2+1
                              average=float(total2/count2)'

                          else:
                              continue

      print 'average='+ str(average)

But, the problem with this is-

If the first element of list, ie ['p','e'] is present, it should do the computation and return the value and it should not enter the else loop. Because when it enters else loop, its computing other values in a11 also.

If i am writing break statement after if loop, then its not computing value for second element of list.

Output should be:

average=2.5

average=3.0

average=2.0

I couldn't exactly what you are doing here, but this is how I would do what you've described.

a10=[['p', 'e'], ['t', 'h','a'],['e', 'a'],['p','e'],['a']]

c10=[['p','e'],['e','h'],['e', 'a']]

a11=[4,5,2,1,3]

# Go through c10 element by element
for c10_elem in c10:
    # List of matching indexes. XXX: if same indexes can't be counted twice, 
    #                                use a set instead of a list
    id = []
    # Look for match in a10 and record it's index
    for i, a10_elem in enumerate(a10):
        if c10_elem == a10_elem:
            id.append(i)

    # We didn't find element level match
    if not id:
        # Look for char level match and record it's index
        for char in c10_elem:
            for i, a10_elem in enumerate(a10):
                if char in a10_elem:
                    id.append(i)

    # find average
    sum = 0
    for i in id:
        sum = sum + a11[i]

    average = sum / len(id)

    print("average {}".format(average))

Output -

$ python stack.py 
average 2.5
average 3.0
average 2.0

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM