将二维数组处理成 Python 中的列表列表

Question

I want to process this 2D array ('more') in order to get a list of lists ('rm'), so that the final result would look like this:

     Month1   Month2   Month3
rm = [[3,       [4],    [5]]    
       2, 
       1, 
       0], 

where the first column of 'more' corresponds to the Month of data collection at the stations. The final output above has 3 lists for the respective 3 Months of data collection, after removing the -999 values. Note that in the original array data for one month could spread to several rows, but they all end up in one single list in the final output.

I am new to python, so I know I am not doing a good job properly framing this. I appreciate your help on this.

rm = []
       Month  Station1 Station2 Station3
more = [[1,    -999,      3,        2], 
        [1,       1,      0,     -999], 
        [2,       4,   -999,     -999],
        [3,    -999,   -999,        5]]

for i in range(0, len(more)):
    rm.append([])
    r = 0
    for j in range(1, len(more[0])):
        col = more[i][0] - 1
        if (more[i][j] > -999):
            rm[r][col].append(more[i][j])
            r = r + 1

print(rm)

with this code I am getting the error "list index out of range" for the line:

rm[r][col].append(more[i][j])

thanks !

Answer 1

You're trying to get the index of an empty list which doesn't have any index values. rm[r][col] will access the 0 index of rm but then try to access the 0 index of the list you appended which is empty and has no indexes. Instead of appending to a list I think a dictionary would be best to hold the values in a more organized and then easier to convert into an array if you have to have it in one. Myself I would keep it in the dictionary that way you could just refer to the value of the month like so data[1] and get the values you need without worrying about indexing.

rm = []
       Month  Station1 Station2 Station3
more = [[1,    -999,      3,        2], 
        [1,       1,      0,     -999], 
        [2,       4,   -999,     -999],
        [3,    -999,   -999,        5]]

data = {}
for i in range(0, len(more)):
    if more[i][0] not in data.keys():
        data[more[i][0]] = []
    for j in range(1, len(more[0])):
        if (more[i][j] > -999):
            data[more[i][0]].append(more[i][j])

rm = [v for k, v in data.items()]

print(rm)

Which gives this output:

[3, 2, 1, 0], [4], [5]]

Answer 2

There are many ways to solve this problem, and you could probably use dict or other data structures to facilitate the problem. Here's a solution that uses a dictionary to hold the lists. This solution works well because you want to associate the items in the list with a month number, but the same month can span multiple sublists.

In python, you can iterate over lists directly without using indices, if you want indices, you can just do for index, item in enumerate(list) rather than using range and len .

Try:

from collections import defaultdict

d = defaultdict(list)
for row in more:
    for val in row[1:]:
        if val != -999:
            # row[0] stores the month value which we want to key on
            d[row[0]].append(j)

rm = list(d.values())

This will still yield the same set of lists in rm as before but you lose the association between which values belong to which month. It is probably better to keep them in dictionary format.