我试图在python中找到音乐片段(来自.xml)中具有相同名称的所有音符的持续时间。我有3个列表:
i
我想将durationsAll列表中的所有持续时间与所有具有相同名称的音符相加。例如对于所有来自notesAll列表的' A等于scale [0],我想要类似于:durationsAll [1] + durationsAll [3] = 1.5 + 1 = 2.5。我需要一个比我的尝试更好的解决方案:
scales = ['A', 'B', 'C'] #scales names
notesAll = ['B', 'A', 'C', 'A', 'C', 'C'] #note names of the piece
durationsAll = [1, 1.5, 1.5, 1, 1, 2] #duration of each note from notesAll list
我想要的结果是:for sc in scales:
for ntPosition, nt in enumerate(notesAll):
dtOfEach = 0
for dtPosition, dt in enumerate(durationsAll):
if sc == nt:
dtPosotion = ntPosition #I guess here is the problem
dtOfEach = dtOfEach + dt
答案 0 :(得分:1)
defaultdict在这里是完美的(假设笔记和笔记的持续时间在相应的列表当然)
from collections import defaultdict
duration_total = defaultdict(int)
for note in list(zip(notesAll, durationsAll)):
duration_total[note[0]] += note[1]
print(duration_total)
{'B':1,'A':2.5,'C':4.5})
我认为A:3.5在你的问题中是一个错字?
的修改
使用Chris的建议更新了代码:
for note, duration in zip(notesAll, durationsAll):
duration_total[note] += duration
答案 1 :(得分:0)
您可以使用索引并按以下方式继续:
c = Clock()
assert(c.minutes == 0 and c.hours == 0)
for i in range(24):
assert(c.hours == i)
for j in range(60):
assert(c.hours == i and c.minutes == j)
c.tic()
assert(c.minutes == 0 and c.hours == 0)
def get_total_duration(note):
return sum(durationsAll[idx] for idx, nt in enumerate(notesAll) if nt == note)
scales = ['A', 'B', 'C'] #scales names
notesAll = ['B', 'A', 'C', 'A', 'C', 'C'] #note names of the piece
durationsAll = [1, 1.5, 1.5, 1, 1, 2]
get_total_duration('A')
答案 2 :(得分:0)
您可以进行间接排序,然后使用groupby
from itertools import groupby
scales = ['A', 'B', 'C'] #scales names
notesAll = ['B', 'A', 'C', 'A', 'C', 'C'] #note names of the piece
durationsAll = [1, 1.5, 1.5, 1, 1, 2] #duration of each note from notesAll list
# do indirect sort
idx = sorted(range(len(notesAll)), key=notesAll.__getitem__)
# group and sum
result = {k: sum(map(durationsAll.__getitem__, grp))
for k, grp in groupby(idx, notesAll.__getitem__)}
# {'A': 2.5, 'B': 1, 'C': 4.5}
答案 3 :(得分:0)
第一笔金额应该是2.5吗?如果是这样的话,就像这条消息末尾的代码应该有效一样。我认为你基本上使用了for循环,然后枚举,但我不确定第二个枚举是什么。
你还有一行:
dtPosotion = ntPosition #I guess here is the problem
dtPosotion是拼写错误吗?或者您是否尝试设置dtPosition = ntPosition。如果是这样,我认为这是不可能的,因为dtPosition和ntPosition是由枚举循环设置的。
scales = ['A', 'B', 'C'] #scales names
notesAll = ['B', 'A', 'C', 'A', 'C', 'C'] #notes names of the piece
durationsAll = [1, 1.5, 1.5, 1, 1, 2] #duration of each note from notesAll list
sums = [0,0,0]
for s in scales:
print "Scale letter:" + str(s)
for i,j in enumerate(notesAll):
if j == s:
sums[scales.index(s)] +=durationsAll[i]
print sums
答案 4 :(得分:0)
可以通过列表理解来完成,利用布尔表现得像整数的想法
In [3]: S = ['A', 'B', 'C'] #scales names
...: N = ['B', 'A', 'C', 'A', 'C', 'C'] #note names of the piece
...: D = [1, 1.5, 1.5, 1, 1, 2] #duration of each note from notesAll list
In [4]: [sum(match*d for match, d in zip((n==s for n in N), D)) for s in S]
Out[4]: [2.5, 1.0, 4.5]
如果你需要字典
In [5]: {s:sum(match*d for match, d in zip((n==s for n in N), D)) for s in S}
Out[5]: {'A': 2.5, 'B': 1.0, 'C': 4.5}
或者,避免一个循环而不导入collections
In [6]: r = {}
In [7]: for n, d in zip(N, D): r[n] = r.get(n, 0) + d
In [8]: r
Out[8]: {'A': 2.5, 'B': 1, 'C': 4.5}
我们访问字典中的数据而不是通过索引,但使用字典.get(key, default)
方法,可以正确初始化我们的值。