Question

好的 - 我确定之前已经回答了这个问题，但我找不到它......

我的问题：我有一个包含此构图的列表列表

0.2 A

0.1 A

0.3 A

0.3 B

0.2 C

0.5 C

我的目标是输出以下内容：

0.6 A

0.3 B

0.7 C

换句话说，我需要将多行数据合并在一起。

这是我正在使用的代码：

unique_percents = []

for line in percents:
    new_percent = float(line[0])
    for inner_line in percents:
        if line[1] == inner_line[1]:
           new_percent += float(inner_line[0])
        else:
            temp = []
            temp.append(new_percent)
            temp.append(line[1])
            unique_percents.append(temp)
            break

我认为它应该可以工作，但它并没有增加百分比，仍然有重复。也许我不明白“休息”是如何运作的？

我还会提出更好的循环结构或算法的建议。谢谢，大卫。

Answer 1

你想使用dict，但是collections.defaultdict可以在这里非常方便，所以你不必担心密钥是否存在于dict中 - 它只是默认为0.0：< / p>

import collections

lines = [[0.2, 'A'], [0.1, 'A'], [0.3, 'A'], [0.3, 'B'], [0.2, 'C'], [0.5, 'C']]
amounts = collections.defaultdict(float)
for amount, letter in lines:
    amounts[letter] += amount

for letter, amount in sorted(amounts.iteritems()):
    print amount, letter

Answer 2

试试这个：

result = {}
for line in percents:
    value, key = line
    result[key] = result.get(key, 0) + float(value)

Answer 3

total = {}
data = [('0.1', 'A'), ('0.2', 'A'), ('.3', 'B'), ('.4', 'B'), ('-10', 'C')]
for amount, key in data:
    total[key] = total.get(key, 0.0) + float(amount)

for key, amount in total.items():
    print key, amount

Answer 4

由于所有字母等级都组合在一起，您可以使用itertools.groupby（如果没有，只需提前对列表进行排序即可）：

data = [
    [0.2, 'A'],
    [0.1, 'A'],
    [0.3, 'A'],
    [0.3, 'B'],
    [0.2, 'C'],
    [0.5, 'C'],
]

from itertools import groupby

summary = dict((k, sum(i[0] for i in items)) 
                for k,items in groupby(data, key=lambda x:x[1]))

print summary

给出：

{'A': 0.60000000000000009, 'C': 0.69999999999999996, 'B': 0.29999999999999999}

Answer 5

如果您有这样的列表列表： [ [0.2, A], [0.1, A], ...]（实际上它看起来像一个元组列表:)

res_dict = {}

for pair in lst:
    letter = pair[1]
    val = pair[0]
    try:
        res_dict[letter] += val
    except KeyError:
        res_dict[letter] = val

res_lst = [(val, letter) for letter, val in res_dict] # note, a list of tuples!

Answer 6

使用collections.defaultdict来计算值（假设d中的文本数据）：

>>> s=collections.defaultdict(float)
>>> for ln in d:
...     v,k=ln.split()
...     s[k] += float(v)
>>> s
defaultdict(<type 'float'>, {'A': 0.60000000000000009, 'C': 0.69999999999999996, 'B': 0.29999999999999999})
>>> ["%s %s" % (v,k) for k,v in s.iteritems()]
['0.6 A', '0.7 C', '0.3 B']
>>>

Answer 7

如果您使用的是Python 3.1或更高版本，则可以使用collections.Counter。另外我建议使用decimal.Decimal而不是浮点数：

# Counter requires python 3.1 and newer
from collections import Counter
from decimal import Decimal

lines = ["0.2 A", "0.1 A", "0.3 A", "0.3 B", "0.2 C", "0.5 C"]
results = Counter()
for line in lines:
    percent, label = line.split()
    results[label] += Decimal(percent)
print(results)

结果是：

计数器（{'C'：十进制（'0.7'），'A'：十进制（'0.6'），'B'：十进制（'0.3'）}）

Answer 8

这很冗长，但有效：

# Python 2.7
lines = """0.2 A
0.1 A
0.3 A
0.3 B
0.2 C
0.5 C"""

lines = lines.split('\n')
#print(lines)
pctg2total = {}
thing2index = {}
index = 0
for line in lines:
    pctg, thing = line.split()
    pctg = float(pctg)
    if thing not in thing2index:
        thing2index[thing] = index
        index = index + 1
        pctg2total[thing] = pctg
    else:
        pctg2total[thing] = pctg2total[thing] + pctg
output = ((pctg2total[thing], thing) for thing in pctg2total)
# Let's sort by the first occurrence.
output = list(sorted(output, key = lambda thing: thing2index[thing[1]]))
print(output)

>>> 
[(0.60000000000000009, 'A'), (0.29999999999999999, 'B'), (0.69999999999999996, 'C')]

Answer 9

letters = {}
for line in open("data", "r"):
    lineStrip = line.strip().split()
    percent = float(lineStrip[0])
    letter = lineStrip[1]
    if letter in letters:
        letters[letter] = percent + letters[letter]
    else:
        letters[letter] = percent

for letter, percent in letters.items():
    print letter, percent

A 0.6
C 0.7
B 0.3

Answer 10

>>> from itertools import groupby, imap
>>> from operator import itemgetter
>>> data = [['0.2', 'A'], ['0.1', 'A'], ['0.3', 'A'], ['0.3', 'B'], ['0.2', 'C'], ['0.5', 'C']]
>>> # data = sorted(data, key=itemgetter(1))
... 
>>> for k, g in groupby(data, key=itemgetter(1)):
...     print sum(imap(float, imap(itemgetter(0), g))), k
... 
0.6 A
0.3 B
0.7 C
>>>

Answer 11

假设我们有这个

data =[(b, float(a)) for a,b in 
    (line.split() for line in
        """
        0.2 A
        0.1 A
        0.3 A
        0.3 B
        0.2 C
        0.5 C""".splitlines()
        if line)]
print data 
# [('A', 0.2), ('A', 0.1), ('A', 0.3), ('B', 0.3), ('C', 0.2), ('C', 0.5)]

你现在可以通过这个和总结

counter = {}
for letter, val in data:
    if letter in counter:
        counter[letter]+=val
    else:
        counter[letter]=val

print counter.items()

或将值组合在一起并使用sum：

from itertools import groupby
# you want the name and the sum of the values
print [(name, sum(value for k,value in grp)) 
    # from each group
    for name, grp in 
    # where the group name of a item `p` is given by `p[0]`
    groupby(sorted(data), key=lambda p:p[0])]

Python：合并计数数据

11 个答案: