Print individual elements of a list in python

Question

I have the following extract from an input file:

Query_7736    1624  SDLA**VY*EMQALRIKPSNVTFSILIKLYGRNKQVSKAIEVLEEMKR*GVQPGMIVYTC  1803
XP_002972017  833   MAEACELMRSLRSLRVSPDTVTFSTLIDGLCKCGQTDEACNVFDDMIAGGYVPNVVTYNV  894
XP_002972017  583   FEQASALFEEMVAKNLQPDVMTFGALIDGLCKAGQVEAARDILDLMGNLGVPPNVVTYNA  642
XP_002972017  653   IEEACQFLEEMVSSGCVPDSITYGSLVYALCRASRTDDALQLVSELKSFGWDPDTVTYNI  712
XP_002972017  905   MERAHAMIESMVDKGVTPDVITYSVLVDAFCKASHVDEALELLHGMASRGCTPNVVTFNS  964
XP_002972017  940   VDEALELLHGMASRGCTPNVVTFNSIIDGLCKSDQSGEAFQMFDDMTKHGLAPDKITYCT  1000
XP_002970953  380   RELASSVYKTMTSHGCVPDVVTLSTMIDGLSKAGRIGAAVRIFKSMEARGLAPNEVVYSA  447
XP_002970953  458   MDCALEMLAQMKKAFCTPDTITYNILIDGLCKSGDVEAARAFFDEMLEAGCKPDVYTYNI  517
XP_002971975  632   LEEARKILERLERENCKADMFAYRVMMDGLCRTGRMSAALELLEAIKQSGTPPRHDIYVA  692
XP_002971975  527   VDDAERLLEEMVASDCSPDVYTYTSLVDGFCKVGRMVEARRVLKRMAKRGCQPNVVTYTA  586
XP_002971975  387   VRDAQEVFKRMIVRGIEPNVVTYNSLIHGFCMTNGVDSALLLMEEMTATGCLPDIITYNT  446
XP_002971975  317   LDEACKLFEKMRENSCEPDVVTFTALMDGLCKGDRLQEAQQVLETMEDRNCTPNVITYSS  376
XP_002961692  489   VRDALGLLEFMIESGLSPDVITFNSVLDGLCKEQRILDAHNVFKRALERGCRPNVVTYST  548
XP_002961692  873   SEQALELLRAMVADGGSPDACNYMTVMDGLFKAGSPEVAAKLLQEMRSRGHSPDLRTYTI  932

I have the following script that searches each line with the 'Query_' looks for an '*' and prints the column under it.

lines = [line.rstrip() for line in open('infile.txt')]
for line in lines:
    data = line.split()
    sequence = data[2]
    if data[0].startswith("Query_"):
        star_indicies = [i for i,c in enumerate(sequence) if c == '*'] 
    else:
        print list(sequence[star_index] for star_index in star_indicies)

The output is generated as follows:

['C', 'E', 'R', 'G']
['S', 'A', 'E', 'L']
['C', 'Q', 'E', 'F']
['H', 'A', 'E', 'R']
['L', 'E', 'H', 'H']
['S', 'S', 'K', 'R']
['L', 'E', 'A', 'A']
['R', 'K', 'E', 'S']
['E', 'R', 'E', 'R']
['Q', 'E', 'K', 'T']
['C', 'K', 'E', 'R']
['L', 'G', 'E', 'R']
['L', 'E', 'R', 'R']
['V', 'E', 'D', 'D']
['L', 'G', 'E', 'A']
['C', 'Q', 'S', 'N']
['L', 'E', 'Q', 'A']

How do I output each individual column to a new line such as this:

C, S, C, H, L, S, L, R, E, Q, C, L, L, V, L, C, L
E, A, Q, A, E, S, E, K, R, E, K, G, E, E, G, Q, E
R, E, E, E, H, K, A, E, E, K, E, E, R, D, E, S, Q
G, L, F, R, H, R, A, S, R, T, R, R, R, D, A, N, A

I can convert this to a string easily, but when I try to print the desired way I am unsuccessful, this is what I was using:

print ("\n".join(map(str,list(sequence[star_index] for star_index in star_indicies))),

Looks like all you need is to [transpose the list before printing](http://stackoverflow.com/a/17037588/6563468) — Logan Reed, Jul 13 '16 at 18:42

xbello · Answer 1 · 2016-07-15T09:29:21.393

If you can have more than one "Query_" per file and in different order:

lines = [line.rstrip().split() for line in open('infile.txt')]

# Load the indexes in one list, the sequences in another
# As shown in http://stackoverflow.com/a/21023591/1688590
indexes, sequences = [], []

for line in lines:
    target = indexes if line[0].startswith("Query_") else sequences
    target.append(line[2])

for pos, char in enumerate(zip(*indexes)):
    # Check if any of the "Query_" sequences has a * in that position
    if "*" in char:
        # Output every char in that position in the other secuences
        print(", ".join([_[pos] for _ in sequences]))

score 0 · Answer 2 · answered Jul 13 '16 at 18:45

Just join each column of your last list with a comma

result = list(sequence[star_index] for star_index in star_indices)
for i in range(len(result[0])):
    print(", ".join([l[i] for l in result]))

Here's what you get

In [2]: result = [['C', 'E', 'R', 'G'],
['S', 'A', 'E', 'L'],
['C', 'Q', 'E', 'F'],
['H', 'A', 'E', 'R'],
['L', 'E', 'H', 'H'],
['S', 'S', 'K', 'R'],
['L', 'E', 'A', 'A'],
['R', 'K', 'E', 'S'],
['E', 'R', 'E', 'R'],
['Q', 'E', 'K', 'T'],
['C', 'K', 'E', 'R'],
['L', 'G', 'E', 'R'],
['L', 'E', 'R', 'R'],
['V', 'E', 'D', 'D'],
['L', 'G', 'E', 'A'],
['C', 'Q', 'S', 'N'],
['L', 'E', 'Q', 'A']]

In [3]: for i in range(len(result[0])):
            print(", ".join([l[i] for l in result]))
   ...:     
C, S, C, H, L, S, L, R, E, Q, C, L, L, V, L, C, L
E, A, Q, A, E, S, E, K, R, E, K, G, E, E, G, Q, E
R, E, E, E, H, K, A, E, E, K, E, E, R, D, E, S, Q
G, L, F, R, H, R, A, S, R, T, R, R, R, D, A, N, A

Or, as others have suggested:

In [9]: for l in zip(*result):
   ...:     print(", ".join(l))
   ...:     
C, S, C, H, L, S, L, R, E, Q, C, L, L, V, L, C, L
E, A, Q, A, E, S, E, K, R, E, K, G, E, E, G, Q, E
R, E, E, E, H, K, A, E, E, K, E, E, R, D, E, S, Q
G, L, F, R, H, R, A, S, R, T, R, R, R, D, A, N, A

That is the output from executing the code in the python interpreter in my answer. Not sure where your brackets are coming from — kdopen, Jul 13 '16 at 19:34
You will find [enumerate](https://docs.python.org/2/library/functions.html#enumerate) cleaner than `range(len(x))` — xbello, Jul 13 '16 at 20:37

score 0 · Answer 3 · answered Jul 13 '16 at 19:02

Given the following

In [7]: list(d2)
Out[7]: 
[['C', 'E', 'R', 'G'],
 ['S', 'A', 'E', 'L'],
 ['C', 'Q', 'E', 'F'],
 ['H', 'A', 'E', 'R'],
 ['L', 'E', 'H', 'H'],
 ['S', 'S', 'K', 'R'],
 ['L', 'E', 'A', 'A'],
 ['R', 'K', 'E', 'S'],
 ['E', 'R', 'E', 'R'],
 ['Q', 'E', 'K', 'T'],
 ['C', 'K', 'E', 'R'],
 ['L', 'G', 'E', 'R'],
 ['L', 'E', 'R', 'R'],
 ['V', 'E', 'D', 'D'],
 ['L', 'G', 'E', 'A'],
 ['C', 'Q', 'S', 'N'],
 ['L', 'E', 'Q', 'A']]

You can use zip() with the unpacking operator (*) like so:

In [8]: '\n'.join([' '.join(l) for l in zip(*d2)])
Out[8]: 'C S C H L S L R E Q C L L V L C L\nE A Q A E S E K R E K G E E G Q E\nR E E E H K A E E K E E R D E S Q\nG L F R H R A S R T R R R D A N A'

Print individual elements of a list in python

3 Answers3