Question

这当然很简单，但现在我已经尝试了几个小时。我想检查一个10x1矩阵中所有值的值，如果它大于它们中的任何值，它应该在大于的元素之前插入。

到目前为止，我已经尝试了以下代码的不同变体但运气不错。我得到的是以下内容：

我尝试过：

col,col1,col2 = np.zeros((10,1)),np.zeros((10,1)),np.zeros((10,1))

for element in col:             
    if (aggdelay>element):                  
        col[n,0] = aggdelay
        col1[n,0] = flight_num
        col2[n,0] = airline_id
        break               

    n +=1
    if (n>10):
        n=0

我得到的输出如下所示：

[[ 157.]
 [   3.]
 [   6.]
 [   6.]
 [   5.]
 [   9.]
 [   0.]
 [   0.]
 [   0.]
 [   0.]]

输入是：

 19790  1256    124.0
19790   1257    157.0
19790   1258    3.0
19790   1264    6.0
19790   1266    6.0
19790   1280    5.0
19790   1282    9.0

预期输出为：

19790   1258    3.0
19790   1280    5.0
19790   1264    6.0
19790   1266    6.0
19790   1282    9.0
19790   1256    124.0
19790   1257    157.0

我实施了David提供的解决方案，但我发现更新＆＃34;矩阵＆＃34;与新元素。这是我现在的解决方案，但我怀疑它没有正确更新。

 #!/usr/bin/python
import sys
import collections
import numpy as np
from operator import itemgetter

result =np.zeros((3,1))
col,col1,col2 = []*10,[]*10,[]*10
col11,col12,col23 = [],[],[]
old_flight_num, old_airline_id = None, None

lines = sys.stdin.readlines()
sumDelay1, num = 0, 1
n = 0
for line in lines:

    line, line = line.strip(), line.split("\t")

    if len(line) !=3:
        continue

    airline_id, flight_num, aggdelay = line

    try:
        aggdelay = float(aggdelay)
        flight_num= int(flight_num)
        airline_id = int(airline_id)
    except ValueError:
        continue

    if (old_airline_id is not None) and (old_airline_id != airline_id):

        res2.sort(key=itemgetter(2))

        print('                                                      ')     
        print('Here come the results for airline ID: ', (old_airline_id))
        print('                                                      ')
        for row in res2:
            print(row)      

        col,col1,col2 = []*10,[]*10,[]*10

        n=0

    if (n<10):
        col.append(airline_id),col1.append(flight_num),col2.append(aggdelay)

    else:   
        res = zip(col,col1,col2)
        res.sort(key=itemgetter(2))

        if (aggdelay>min(col2)):
            res.remove(res[0])
            col11.append(airline_id), col12.append(flight_num), col23.append(aggdelay)
            res1 = zip(col11,col12,col23)
            res2=res+res1

            res2.sort(key=itemgetter(2))
    col11,col12,col23 = [],[],[]    
    n += 1

    old_airline_id = airline_id

if (old_airline_id is not None):

    res2.sort(key=itemgetter(2))
    print('                                                      ')     
    print('Here come the results for airline ID: ', (old_airline_id))
    print('                                                      ')
    for row in res2:
        print(row)

我非常感谢对此的一些指导。谢谢！

Answer 1

这可能会有所帮助，但我不得不猜测你对输出的期望以及你如何处理输入（假设在3列中给出了航空公司ID，航班号，延误）。

import numpy as np
from operator import itemgetter

ids = [19790,19790,19790,19790,19790,19790,19790,19790]
flight_nums = [1256,1257,1258,1264,1266,1280,1282]
agg_delays = [124.0,157.0,3.,6.,6.,5.,9.]
m = zip(agg_delays,flight_nums,ids)  
m.sort(key=itemgetter(0),reverse=True)  # sorts the zipped list by delay, decreasing

matrix = np.array(m, np.float32)  # dumps the sorted list in to your matrix, an ndarray

这会为您提供一个包含3列的ndarray对象，第一列是您的“延迟”，然后是航班号，然后是航空公司ID，它按第一列排序。

如果您只对前N个感兴趣，那么只需使用上面的内容构建整个矩阵并对其进行切片：

# return the top 10, or however many:
matrix = matrix[:10]

不使用list.sort方法和切片，您可以按相反的顺序对ndarray对象进行排序：

m = zip(agg_delays,flight_nums,ids)
matrix = np.array(m, np.float32)
matrix.sort(0)
# return the top 10, or however many:
matrix = matrix[::-1][:10]

如何在Python中检查矢量/矩阵中每个元素的值

1 个答案: