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[英]Applying functions/methods within each sublist, not to the list of sublists as a whole in python
[英]Applying arithmetic functions to elements of sublists where the first element of each sublist is the same
这是我的问题。
我有一个列表列表,如下所示:
linesort=[
['Me', 1, 596],
['Mine', 1, 551],
['Myself', 1, 533],
['Myself', 1, 624],
['Myself', 1, 656],
['Myself', 1, 928],
['Theirs', 1, 720],
['Theirs', 1, 1921],
['Them', 1, 716],
['Themselves', 1, 527]
]
每个子列表代表参与者对单词进行正确或错误分类的时间(第二个值)和响应时间(第三个值)。 我想做的是返回另一个列表列表,该列表包含单词,每个列表中第二个值的总和以及第三个值的平均值。
本质上,我需要比较每个子列表的第一个元素,如果它们相等,则计算第二个元素的和与第三个元素的平均值。
虽然我能够手动执行此操作(即手动分配和创建变量),但我在循环中执行此操作的尝试均失败了。 鉴于我有两个带有此类数据的非常大的文本文件,我希望采用编程方式。
一些要点可能是有用的:我预先知道每个测试中使用的单词,但我不知道它们会出现在哪里(即使它们出现在任何一组刺激中)。 谁能帮我这个忙吗?
我在Ubuntu 10.04上使用Python 2.6.5。
不是美丽的一个,但:
from collections import defaultdict
linesort = [['Me', 1, 596], ['Mine', 1, 551], ['Myself', 1, 533], ['Myself', 1, 624],
['Myself', 1, 656], ['Myself', 1, 928], ['Theirs', 1, 720],
['Theirs', 1, 1921], ['Them', 1, 716], ['Themselves', 1, 527]]
d = defaultdict(list)
for line in linesort:
d[line[0]].append(line[1:])
output = {}
for x,val in d.items():
svals = [y[1] for y in val]
output[x] = [sum([y[0] for y in val]), sum(svals) / len(svals)] # need to be modified if you need float value
print output
>>> {'Mine': [1, 551], 'Theirs': [2, 1320], 'Me': [1, 596], 'Them': [1, 716], 'Themselves': [1, 527], 'Myself': [4, 685]}
或使用groupby(请注意,这不是最有效的,并且需要对包含初始数据的列表进行排序):
from itertools import groupby
res = {}
for key, gen in groupby(sorted(linesort), key=lambda x: x[0]):
val = list(gen)
svals = [y[2] for y in val]
res[key] = [sum([y[1] for y in val]), sum(svals) / float(len(svals))]
但是我以前的所有示例都为您返回了字典,因此,如果您想获取列表,则只需要稍微修改一下代码即可:
from itertools import groupby
res = []
for key, gen in groupby(sorted(linesort), key=lambda x: x[0]):
val = list(gen)
svals = [y[2] for y in val]
res.append([key, sum([y[1] for y in val]), sum(svals) / float(len(svals))])
print res
>>> [['Me', 1, 596.0], ['Mine', 1, 551.0], ['Myself', 4, 685.25], ['Theirs', 2, 1320.5], ['Them', 1, 716.0], ['Themselves', 1, 527.0]]
我的详细解决方案
#!/usr/bin/env python
import collections
linesort=[['Me', 1, 596], ['Mine', 1, 551], ['Myself', 1, 533], ['Myself', 1, 624],
['Myself', 1, 656], ['Myself', 1, 928],['Theirs', 1, 720], ['Theirs', 1, 1921],
['Them', 1, 716], ['Themselves', 1, 527]]
new=[]
d=collections.defaultdict(list)
for i in linesort:
d[i[0]].append(i[1:])
for k,v in d.iteritems():
s=sum([i[0] for i in v])
avg=sum([i[1] for i in v]) / len(v)
new.append([k,s,avg])
for i in new: print i
输出:
['Me', 1, 596]
['Myself', 4, 685]
['Theirs', 2, 1320]
['Mine', 1, 551]
['Themselves', 1, 527]
['Them', 1, 716]
这是我的简单解决方案:
#!/usr/bin/python
linesort=[['Me', 1, 596], ['Mine', 1, 551], ['Myself', 1, 533], ['Myself', 1, 624], ['Myself', 1, 656], ['Myself', 1, 928], ['Theirs', 1, 720], ['Theirs', 1, 1921], ['Them', 1, 716], ['Themselves', 1, 527]]
cnts = {};
sums = {};
# here we count occurrences of each word (cnts),
# and we compute the the sum of second elements of each input list
for list in linesort:
cnts[list[0]] = cnts.get(list[0], 0) + 1;
sums[list[0]] = sums.get(list[0], 0) + list[1];
# now that we know the occurrences for each work we can compute
# the averages of the third elements of each input list
avgs = {};
for list in linesort:
avgs[list[0]] = avgs.get(list[0], 0) + list[2] / cnts[list[0]];
# we build the result as a list of lists
result = [];
for word in avgs:
result.append([word, sums[word], avgs[word]]);
print result;
输出为:
[['Me', 1, 596], ['Myself', 4, 685], ['Theirs', 2, 1320], ['Mine', 1, 551], ['Themselves', 1, 527], ['Them', 1, 716]]
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