Question

我已向此发布了一个问题similar，但我仍然遇到过一些问题。

我有一个看起来像这样的元组列表：

[(1, 0.5, 'min'),
 (2, 3, 'NA'),
 (3, 6, 'NA'),
 (4, 40, 'NA'),
 (5, 90, 'NA'),
 (6, 130.8, 'max'),
 (7, 129, 'NA'),
 (8, 111, 'NA'),
 (9, 8, 'NA'),
 (10, 9, 'NA'),
 (11, 0.01, 'min'),
 (12, 9, 'NA'),
 (13, 40, 'NA'),
 (14, 90, 'NA'),
 (15, 130.1, 'max'),
 (16, 112, 'NA'),
 (17, 108, 'NA'),
 (18, 90, 'NA'),
 (19, 77, 'NA'),
 (20, 68, 'NA'),
 (21, 0.9, 'min'),
 (22, 8, 'NA'),
 (23, 40, 'NA'),
 (24, 90, 'NA'),
 (25, 92, 'NA'),
 (26, 130.4, 'max')]

我正在进行实验，每个实验只有一个＆＃34; min＆＃34;和一个＆＃34; max＆＃34;值。我想总结第一个元素中的元素，最多只有一个＆＃34; min＆＃34;和＆＃34;最大＆＃34;例如，这个小数据集有3个实验，因为有3分钟和3个最大值。输出看起来像：

exp = [1+2+3+4+5+6+7+8+9+10, 11+12+13+14+15+16+17+18+19+20, 21+22+23+24+25+26]

我还想跟踪添加到列表中的值，以便我也有这个输出：

exp_values = [[1,2,3,4,5,6,7,8,9,10], [11,12,13,14,15,16,17,18,19, 20], [21, 22, 23, 24, 25, 26]]

我在尝试入门时遇到了麻烦，到目前为止只有一个大致的想法：

times = []
sum_
for item in tup_list:
    if item[2] != "min":
       sum_ += item[0]
       times.append(sum_)

Answer 1

Segmentation fault (core dumped)

你走了。我不认为python代码需要在这里描述。

vals = [(1, 0.5, 'min'),
 (2, 3, 'NA'),
 (3, 6, 'NA'),
 (4, 40, 'NA'),
 (5, 90, 'NA'),
 (6, 130.8, 'max'),
 (7, 129, 'NA'),
 (8, 111, 'NA'),
 (9, 8, 'NA'),
 (10, 9, 'NA'),
 (11, 0.01, 'min'),
 (12, 9, 'NA'),
 (13, 40, 'NA'),
 (14, 90, 'NA'),
 (15, 130.1, 'max'),
 (16, 112, 'NA'),
 (17, 108, 'NA'),
 (18, 90, 'NA'),
 (19, 77, 'NA'),
 (20, 68, 'NA'),
 (21, 0.9, 'min'),
 (22, 8, 'NA'),
 (23, 40, 'NA'),
 (24, 90, 'NA'),
 (25, 92, 'NA'),
 (26, 130.4, 'max')]
it = iter(vals)

out = []
ap = [next(it)[0]]
for e,_,state in it:
    if state == 'min':
        out.append(ap)
        ap = []
    ap += [e]
out.append(ap)
print(out)

Answer 2

这是一种处理它的简单方法：

l = []

for i in tup_list:
    if 'min' in i:
        try:
            l.append(temp)
            temp = []
        except:
            temp = []
    temp.append(i[0])

if len(temp) > 0:
    l.append(temp)

print l

[[1,2,3,4,5,6,7,8,9,10], [11,12,13,14,15,16,17,18,19, 20], [21, 22, 23, 24, 25, 26]]

这将获得分数列表，例如变量exp_values。要获得这些值的总和，只需对列表求和：

sums = list(map(sum, l))

print sums

[55, 155, 141]

Answer 3

这是一个解决方案，它跟踪作为每个实验一部分的指数，然后计算最后的总和。

tup_list = [(1, 0.5, 'min'),
 (2, 3, 'NA'),
 (3, 6, 'NA'),
 (4, 40, 'NA'),
 (5, 90, 'NA'),
 (6, 130.8, 'max'),
 (7, 129, 'NA'),
 (8, 111, 'NA'),
 (9, 8, 'NA'),
 (10, 9, 'NA'),
 (11, 0.01, 'min'),
 (12, 9, 'NA'),
 (13, 40, 'NA'),
 (14, 90, 'NA'),
 (15, 130.1, 'max'),
 (16, 112, 'NA'),
 (17, 108, 'NA'),
 (18, 90, 'NA'),
 (19, 77, 'NA'),
 (20, 68, 'NA'),
 (21, 0.9, 'min'),
 (22, 8, 'NA'),
 (23, 40, 'NA'),
 (24, 90, 'NA'),
 (25, 92, 'NA'),
 (26, 130.4, 'max')]

all_experiment_data = []
current_experiment_data = []
for item in tup_list:
    index, _, point_type = item
    if point_type=="min" and current_experiment_data: #Starting a new experiment, flush the old one
        all_experiment_data.append(current_experiment_data)
        current_experiment_data = []
    current_experiment_data.append(index)

#Flush the last experiment
all_experiment_data.append(current_experiment_data)

all_experiment_sums = [sum(experiment_indices) for experiment_indices in 
all_experiment_data]

print("Indices in each of the experiments:")
print(all_experiment_data)
print("Sums of indices for experiments:")
print(all_experiment_sums)

这会产生：

Indices in each of the experiments:
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], [21, 22, 23, 24, 25, 26]]
Sums of indices for experiments:
[55, 155, 141]

Answer 4

对于此序列，您需要跟踪min和max遇到的次数。一旦新的min进入，它就是新序列的开始。所以我使用inSeq作为序列的跟踪器。让我知道它是否有帮助：

times = []
sum_ = 0
inSeq = 0
for item in tup_list:
    if(item[2] == 'min' or item[2] == 'max'):
        inSeq += 1
    if(inSeq == 3): #start of new sequence when it hits new min
        times.append(sum_)
        sum_ = item[0] 
        inSeq = 1 
    else:  
        sum_ += item[0]

Python以间隔计算列表中的值

4 个答案: