Question

我有以下数据，我想找到＆＃39; a＆＃39;中的唯一值。并汇总来自＆＃39;＆＃39;＆＃39; c＆＃39;的相应指数的数据。有关最佳方法的任何想法吗？我不知道从哪里开始。

a = ['x', 'y', 'z', 'z', 'x', 'w']
b = [  1,   4,   5,   7,   9,  5]
c = [  3,   6,   7,   8,   9,  7]

处理完毕后，

a = ['x', 'y',  'z',  'w']
b = [ 10,   4,   12,   5 ]
c = [ 12,   6,   15,   7 ]

Answer 1

可以使用OrderedDict执行此类操作，因为您需要保持相同的顺序：

from collections import OrderedDict

a = ['x', 'y', 'z', 'z', 'x', 'w']
b = [  1,   4,   5,   7,   9,  5]
c = [  3,   6,   7,   8,   9,  7]
b_data = OrderedDict()
c_data = OrderedDict()

for letter, b_value, c_value in zip(a, b, c):
    if letter in b_data:
        b_data[letter] += b_value
        c_data[letter] += c_value
    else:
        b_data[letter] = b_value
        c_data[letter] = c_value

a = b_data.keys()
b = b_data.values()
c = c_data.values()


print(a)
print(b)
print(c)

输出：

['x', 'y', 'z', 'w']
[10, 4, 12, 5]
[12, 6, 15, 7]

Answer 2

使用collections.defaultdict：

from collections import defaultdict

a = ['x', 'y', 'z', 'z', 'x', 'w']
b = [  1,   4,   5,   7,   9,  5]
c = [  3,   6,   7,   8,   9,  7]

b_unique = collections.defaultdict(int)
c_unique = collections.defaultdict(int)
for k, bv, cv in zip(a,b,c):
    b_unique[k] += bv
    c_unique[k] += cv

Answer 3

使用熊猫：

import pandas as pd

# To keep original order of values.
a_ordered = [val for idx, val in enumerate(a) if val not in a[:idx]]
# >>> a_ordered
# OUT: ['x', 'y', 'z', 'w']

df = pd.DataFrame({'a': a, 'b': b, 'c': c}).groupby('a').sum().T[a_ordered]
a = df.columns.tolist()
b, c = df.values.tolist()

>>> a
['x', 'y', 'z', 'w']

>>> b
[10, 4, 12, 5]

>>> c
[12, 6, 15, 7]

Answer 4

感觉就像玩一些小码 - 高尔夫：

>>> from collections import OrderedDict
>>> od = OrderedDict()
>>> for t in zip(a,zip(b,c)):
...   od[t[0]] = [i + x for i,x in zip(od.get(t[0],[0,0]), t[1])]
... 
>>> od
OrderedDict([('x', [10, 12]), ('y', [4, 6]), ('z', [12, 15]), ('w', [5, 7])])
>>> a = list(od.keys())
>>> b,c = map(list,zip(*od.values()))
>>> a
['x', 'y', 'z', 'w']
>>> b
[10, 4, 12, 5]
>>> c
[12, 6, 15, 7]
>>>

Answer 5

这是一个基本的方法

d1 = defaultdict(int)
d2 = defaultdict(int)
for x,y,z in zip(a,b,c):
    d1[x] += y
    d2[x] += z

an=list()
bn=list()
cn=list()

for k in sorted(d1.keys(), key=lambda x:a.index(x)):
    an.append(k)
    bn.append(d1[k])
    cn.append(d2[k])

[an,bn,cn]


[['x', 'y', 'z', 'w'], [10, 4, 12, 5], [12, 6, 15, 7]]

在python中连接列表中的元素

5 个答案: