遍历列表字典并返回相同索引处的字符

Question

Working on printing a vertical column from a matrix passed as a string.

I've created a dictionary and assigned each row of the matrix as a value in the dict then bracketed to create a dictionary of lists.

Would like to iterate through each key in the dict and append the value of the given index (eg if value is 'ab c', return 'a' for 1, ' ' for 2...) but all I keep getting is:

[['a b c '], ['a b c '], ['a b c ']]

Or variations on this when I fiddle with it. It never seems to get past row 1, although each value is clearly a different row in the matrix.

Appreciate any help.

def column (str, index):
    output = []
    li = str.split("\n")
    row_dict = {
            1: [li[0]],
            2: [li[1]],
            3: [li[2]] 
            }
    for key in row_dict:
        output.append(row_dict[index])
    return output

str = "a b c \n d e f \n g h i"
column(str, 1)

Answer 1

As far as I can see, the only issue with your code is that you are appending the dictionary value (which is a row, and not the actual value, for example it is getting the row at key = 'index', rather than the value at location 'index' in each of the dictionaries) to output, when you want to assign a particular value from each row... this is what you should be doing:

for key in row_dict:
    output.append(row_dict[key].split()[index])
    print (row_dict[key].split()[index])

For index=1, this will print:

b
e
h

This does three things in one statement:

1. gets the string stored at key='key' from dictionary
2. splits your string into individual characters (so you can extract them more easily)
3. Gets the character/word at the index specified by your parameters.

Answer 2

First, split on "\n " because you seem to have a whitespace after each newline.

Getting the nth item of each row is pretty straightforward if you use list comprehensions, eg [row[index] for row in s.split("\n ")] .

Altogether:

>>> def column (s, index):
    return [row[index] for row in s.split("\n ")]

>>> s = "a b c \n d e f \n g h i"
>>> column(s, 1)
[' ', ' ', ' ']

Or, if you want it to be 1-indexed (like in the example in the question) instead of 0-indexed:

>>> def column (s, index):
    return [row[index-1] for row in s.split("\n ")]

>>> s = "a b c \n d e f \n g h i"
>>> column(s, 1)
['a', 'd', 'g']

Answer 3

You probably forgot to mention the key on which you were iterating. Your function should be something like:

def column(str, index):
    output = []
    li = str.split("\n")
    row_dict = {
            1: li[0].lstrip().split(' '),
            2: li[1].strip().split(' '),
            3: li[2].strip().split(' ')
    }
    for key in row_dict:
        output.append(row_dict[key][index])
    return output

Also, note that you were adding extra [] to the values of the row_dict . Finally, iterable objects in Python start from a 0th index, so you would call your function like column("ab c \ndef \nghi", 0) .

Hope it helps.