Сумма столбца с условием

Question

У меня есть данные в текстовом файле. пример.

A B C D E F 
10 0 0.9775 39.3304 0.9311 60.5601 
10 1 0.9802 32.3287 0.9433 56.1201 
10 2 0.9816 39.9759 0.9446 54.0428 
10 3 0.9737 37.8779 0.9419 56.3865 
10 4 0.9798 34.9152 0.905 69.0879 
10 5 0.9803 50.057 0.9201 64.6289 
10 6 0.9805 39.1062 0.9093 68.4061 
10 7 0.9781 33.8874 0.9327 60.7631 
10 8 0.9802 32.5734 0.9376 60.9165 
10 9 0.9798 32.3466 0.94 54.7645 
11 0 0.9749 40.2712 0.9042 71.2873 
11 1 0.9755 35.6546 0.9195 63.7436 
11 2 0.9766 36.753 0.9507 51.7864 
11 3 0.9779 35.6485 0.9371 59.2483 
11 4 0.9803 35.2712 0.8833 79.0257 
11 5 0.981 46.5462 0.9156 66.6951 
11 6 0.9809 41.8181 0.8642 83.7533 
11 7 0.9749 36.7484 0.9259 62.36 
11 8 0.9736 36.8859 0.9395 58.1538 
11 9 0.98 32.4069 0.9255 61.202 
12 0 0.9812 37.2547 0.9121 68.1347 
12 1 0.9808 31.4568 0.9372 55.9992 
12 2 0.9813 36.5316 0.9497 53.1687 
12 3 0.9803 33.1063 0.9051 69.8894 
12 4 0.9786 35.0318 0.8968 72.9963 
12 5 0.9756 63.441 0.9091 69.9482 
12 6 0.9804 39.1602 0.9156 65.2399 
12 7 0.976 35.5875 0.9248 62.6284 
12 8 0.9779 33.7774 0.9416 56.3755 
12 9 0.9804 32.0849 0.9401 55.2871 

Я хочу сумму столбца C. При условии, что. Столбец A имеет уникальное значение (10 строк). Пожалуйста, посоветуй мне.

Answer 1

with open('data.txt') as f: 
    next(f) 
    d=dict() 
    for x in f: 
     if x.split()[0] not in d: 
      d[x.split()[0]]=float(x.split()[2]) 
     else: 
      d[x.split()[0]]+=float(x.split()[2]) 

выход:

{'11': 9.7756, '10': 9.791699999999999, '12': 9.7925} 

Answer 2

>>> L=map(str.split, """10 0 0.9775 39.3304 0.9311 60.5601 
... 10 1 0.9802 32.3287 0.9433 56.1201 
... 10 2 0.9816 39.9759 0.9446 54.0428 
... 10 3 0.9737 37.8779 0.9419 56.3865 
... 10 4 0.9798 34.9152 0.905 69.0879 
... 10 5 0.9803 50.057 0.9201 64.6289 
... 10 6 0.9805 39.1062 0.9093 68.4061 
... 10 7 0.9781 33.8874 0.9327 60.7631 
... 10 8 0.9802 32.5734 0.9376 60.9165 
... 10 9 0.9798 32.3466 0.94 54.7645 
... 11 0 0.9749 40.2712 0.9042 71.2873 
... 11 1 0.9755 35.6546 0.9195 63.7436 
... 11 2 0.9766 36.753 0.9507 51.7864 
... 11 3 0.9779 35.6485 0.9371 59.2483 
... 11 4 0.9803 35.2712 0.8833 79.0257 
... 11 5 0.981 46.5462 0.9156 66.6951 
... 11 6 0.9809 41.8181 0.8642 83.7533 
... 11 7 0.9749 36.7484 0.9259 62.36 
... 11 8 0.9736 36.8859 0.9395 58.1538 
... 11 9 0.98 32.4069 0.9255 61.202 
... 12 0 0.9812 37.2547 0.9121 68.1347 
... 12 1 0.9808 31.4568 0.9372 55.9992 
... 12 2 0.9813 36.5316 0.9497 53.1687 
... 12 3 0.9803 33.1063 0.9051 69.8894 
... 12 4 0.9786 35.0318 0.8968 72.9963 
... 12 5 0.9756 63.441 0.9091 69.9482 
... 12 6 0.9804 39.1602 0.9156 65.2399 
... 12 7 0.976 35.5875 0.9248 62.6284 
... 12 8 0.9779 33.7774 0.9416 56.3755 
... 12 9 0.9804 32.0849 0.9401 55.2871""".split("\n")) 
>>> from collections import defaultdict 
>>> D = defaultdict(float) 
>>> for a,b,c,d,e,f in L: 
...  D[a] += float(c) 
... 
>>> D 
defaultdict(<type 'float'>, {'11': 9.7756, '10': 9.791699999999999, '12': 9.7925}) 
>>> dict(D.items()) 
{'11': 9.7756, '10': 9.791699999999999, '12': 9.7925} 

Answer 3

import csv 
with open("file.txt","rU") as f: 
    reader = csv.reader(f) 
    # read header 
    reader.next() 
    # summarize 
    a_values = [] 
    sum = 0 
    for row in reader: 
     if row[0] not in a_values: 
      a_values.append(row[0]) 
      sum += float(row[2]) 

Answer 4

Для удовольствия

#!/usr/bin/env ksh 

while <file; do 
    ((a[$1]+=$3)) 
done 
print -C a 

выход

([10]=9.7917 [11]=9.7756 [12]=9.7925) 

Требуется недокументированная опция компиляции FILESCAN.

Answer 5

Если вы хотите сумму сгруппированных по значению:

awk '{sums[$1] += $3} END {for (sum in sums) print sum, sums[sum]}' inputfile