Displaying elements of a list, in columns

I have a list of strings. I need to list them in rows and columns. Each row should not have more than "cols" number of values. Each of the values in a given row should be "step" away from the previous value. The values should appear only once in the output. Here is what I have. Any better way to write this code?

cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



vlen = len(vlist)

start = 0

while start < vlen and start < step:

    num = 0

    for idx in range(start, vlen, step):

        if num < cols:

            print(vlist[idx], end=", ")

            num += 1

    print("n")

    start += 1

edited yesterday

200_success

130k17153419

asked yesterday

RebornCodeLover

262

New contributor

$begingroup$
Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
$endgroup$
– greybeard
yesterday

add a comment |

cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



vlen = len(vlist)

start = 0

while start < vlen and start < step:

    num = 0

    for idx in range(start, vlen, step):

        if num < cols:

            print(vlist[idx], end=", ")

            num += 1

    print("n")

    start += 1

edited yesterday

200_success

130k17153419

asked yesterday

RebornCodeLover

262

New contributor

$begingroup$
Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
$endgroup$
– greybeard
yesterday

add a comment |

cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



vlen = len(vlist)

start = 0

while start < vlen and start < step:

    num = 0

    for idx in range(start, vlen, step):

        if num < cols:

            print(vlist[idx], end=", ")

            num += 1

    print("n")

    start += 1

edited yesterday

200_success

130k17153419

asked yesterday

RebornCodeLover

262

New contributor

cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



vlen = len(vlist)

start = 0

while start < vlen and start < step:

    num = 0

    for idx in range(start, vlen, step):

        if num < cols:

            print(vlist[idx], end=", ")

            num += 1

    print("n")

    start += 1

python python-3.x formatting

edited yesterday

200_success

130k17153419

asked yesterday

RebornCodeLover

262

New contributor

edited yesterday

200_success

130k17153419

asked yesterday

RebornCodeLover

262

New contributor

edited yesterday

200_success

130k17153419

edited yesterday

200_success

130k17153419

edited yesterday

200_success

130k17153419

asked yesterday

RebornCodeLover

262

New contributor

asked yesterday

RebornCodeLover

262

asked yesterday

RebornCodeLover

262

New contributor

RebornCodeLover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$begingroup$
Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
$endgroup$
– greybeard
yesterday

add a comment |

$begingroup$
Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
$endgroup$
– greybeard
yesterday

Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.

– greybeard
yesterday

add a comment |

3 Answers
3

active

oldest

votes

Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.

from itertools import islice



cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



for row in range(step):

    print(', '.join(islice(vlist, row, cols * step, step)), end=", nn")

edited yesterday

answered yesterday

200_success

130k17153419

$begingroup$
Thanks for the compact code.
$endgroup$
– RebornCodeLover
yesterday

add a comment |

The code can be made more understandable by:

introducing row and column indices

replace the while loop with a for loop

calculate the index for vlist from the values of the row/col indices

This reduces the number of help variables needed and could result in something like this:

vlist = ["Value" + str(i+1) for i in range(100)]



cols = 4

rows = 10

for row_idx in range(rows):

    for col_idx in range(cols):



        idx = row_idx + rows * col_idx

        print(vlist[idx], end=", ")



    print("n")

answered yesterday

Jan Kuiken

87838

$begingroup$
Thanks Jan. I used this approach as well. But your code looks cleaner :)
$endgroup$
– RebornCodeLover
yesterday

add a comment |

It can be solved by using the slice object.

From Python documentation:

slice - An object usually containing a portion of a sequence. A slice is created using the subscript notation, with colons between numbers
when several are given, such as in variable_name[1:3:5]. The bracket
(subscript) notation uses slice objects internally.

cols = 4 

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



end = step * cols

for start in range(step):

    print(', '.join(vlist[start:end:step]))

Output

Value1, Value11, Value21, Value31

Value2, Value12, Value22, Value32

Value3, Value13, Value23, Value33

Value4, Value14, Value24, Value34

Value5, Value15, Value25, Value35

Value6, Value16, Value26, Value36

Value7, Value17, Value27, Value37

Value8, Value18, Value28, Value38

Value9, Value19, Value29, Value39

Value10, Value20, Value30, Value40

answered 9 hours ago

MiniMax

1363

New contributor

$begingroup$
Thank you @MinMax. 200_success also gave the same answer earlier.
$endgroup$
– RebornCodeLover
9 hours ago

$begingroup$
@RebornCodeLover It is different. @200_success uses the islice from itertools. I use built-in slice. But yes, they are similar in other ways.
$endgroup$
– MiniMax
9 hours ago

$begingroup$
True. Yes I like it. Thanks
$endgroup$
– RebornCodeLover
9 hours ago

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.

from itertools import islice



cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



for row in range(step):

    print(', '.join(islice(vlist, row, cols * step, step)), end=", nn")

edited yesterday

answered yesterday

200_success

130k17153419

$begingroup$
Thanks for the compact code.
$endgroup$
– RebornCodeLover
yesterday

add a comment |

Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.

from itertools import islice



cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



for row in range(step):

    print(', '.join(islice(vlist, row, cols * step, step)), end=", nn")

edited yesterday

answered yesterday

200_success

130k17153419

$begingroup$
Thanks for the compact code.
$endgroup$
– RebornCodeLover
yesterday

add a comment |

Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.

from itertools import islice



cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



for row in range(step):

    print(', '.join(islice(vlist, row, cols * step, step)), end=", nn")

edited yesterday

answered yesterday

200_success

130k17153419

Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.

from itertools import islice



cols = 4

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



for row in range(step):

    print(', '.join(islice(vlist, row, cols * step, step)), end=", nn")

edited yesterday

answered yesterday

200_success

130k17153419

edited yesterday

answered yesterday

200_success

130k17153419

answered yesterday

200_success

130k17153419

answered yesterday

200_success

130k17153419

$begingroup$
Thanks for the compact code.
$endgroup$
– RebornCodeLover
yesterday

add a comment |

$begingroup$
Thanks for the compact code.
$endgroup$
– RebornCodeLover
yesterday

Thanks for the compact code.

– RebornCodeLover
yesterday

add a comment |

The code can be made more understandable by:

introducing row and column indices

replace the while loop with a for loop

calculate the index for vlist from the values of the row/col indices

This reduces the number of help variables needed and could result in something like this:

vlist = ["Value" + str(i+1) for i in range(100)]



cols = 4

rows = 10

for row_idx in range(rows):

    for col_idx in range(cols):



        idx = row_idx + rows * col_idx

        print(vlist[idx], end=", ")



    print("n")

answered yesterday

Jan Kuiken

87838

$begingroup$
Thanks Jan. I used this approach as well. But your code looks cleaner :)
$endgroup$
– RebornCodeLover
yesterday

add a comment |

The code can be made more understandable by:

introducing row and column indices

replace the while loop with a for loop

calculate the index for vlist from the values of the row/col indices

This reduces the number of help variables needed and could result in something like this:

vlist = ["Value" + str(i+1) for i in range(100)]



cols = 4

rows = 10

for row_idx in range(rows):

    for col_idx in range(cols):



        idx = row_idx + rows * col_idx

        print(vlist[idx], end=", ")



    print("n")

answered yesterday

Jan Kuiken

87838

$begingroup$
Thanks Jan. I used this approach as well. But your code looks cleaner :)
$endgroup$
– RebornCodeLover
yesterday

add a comment |

The code can be made more understandable by:

introducing row and column indices

replace the while loop with a for loop

calculate the index for vlist from the values of the row/col indices

This reduces the number of help variables needed and could result in something like this:

vlist = ["Value" + str(i+1) for i in range(100)]



cols = 4

rows = 10

for row_idx in range(rows):

    for col_idx in range(cols):



        idx = row_idx + rows * col_idx

        print(vlist[idx], end=", ")



    print("n")

answered yesterday

Jan Kuiken

87838

The code can be made more understandable by:

introducing row and column indices

replace the while loop with a for loop

calculate the index for vlist from the values of the row/col indices

This reduces the number of help variables needed and could result in something like this:

vlist = ["Value" + str(i+1) for i in range(100)]



cols = 4

rows = 10

for row_idx in range(rows):

    for col_idx in range(cols):



        idx = row_idx + rows * col_idx

        print(vlist[idx], end=", ")



    print("n")

answered yesterday

Jan Kuiken

87838

answered yesterday

Jan Kuiken

87838

answered yesterday

Jan Kuiken

87838

answered yesterday

Jan Kuiken

87838

$begingroup$
Thanks Jan. I used this approach as well. But your code looks cleaner :)
$endgroup$
– RebornCodeLover
yesterday

add a comment |

$begingroup$
Thanks Jan. I used this approach as well. But your code looks cleaner :)
$endgroup$
– RebornCodeLover
yesterday

Thanks Jan. I used this approach as well. But your code looks cleaner :)

– RebornCodeLover
yesterday

add a comment |

It can be solved by using the slice object.

From Python documentation:

slice - An object usually containing a portion of a sequence. A slice is created using the subscript notation, with colons between numbers
when several are given, such as in variable_name[1:3:5]. The bracket
(subscript) notation uses slice objects internally.

cols = 4 

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



end = step * cols

for start in range(step):

    print(', '.join(vlist[start:end:step]))

Output

Value1, Value11, Value21, Value31

Value2, Value12, Value22, Value32

Value3, Value13, Value23, Value33

Value4, Value14, Value24, Value34

Value5, Value15, Value25, Value35

Value6, Value16, Value26, Value36

Value7, Value17, Value27, Value37

Value8, Value18, Value28, Value38

Value9, Value19, Value29, Value39

Value10, Value20, Value30, Value40

answered 9 hours ago

MiniMax

1363

New contributor

$begingroup$
Thank you @MinMax. 200_success also gave the same answer earlier.
$endgroup$
– RebornCodeLover
9 hours ago

$begingroup$
@RebornCodeLover It is different. @200_success uses the islice from itertools. I use built-in slice. But yes, they are similar in other ways.
$endgroup$
– MiniMax
9 hours ago

$begingroup$
True. Yes I like it. Thanks
$endgroup$
– RebornCodeLover
9 hours ago

add a comment |

It can be solved by using the slice object.

From Python documentation:

slice - An object usually containing a portion of a sequence. A slice is created using the subscript notation, with colons between numbers
when several are given, such as in variable_name[1:3:5]. The bracket
(subscript) notation uses slice objects internally.

cols = 4 

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



end = step * cols

for start in range(step):

    print(', '.join(vlist[start:end:step]))

Output

Value1, Value11, Value21, Value31

Value2, Value12, Value22, Value32

Value3, Value13, Value23, Value33

Value4, Value14, Value24, Value34

Value5, Value15, Value25, Value35

Value6, Value16, Value26, Value36

Value7, Value17, Value27, Value37

Value8, Value18, Value28, Value38

Value9, Value19, Value29, Value39

Value10, Value20, Value30, Value40

answered 9 hours ago

MiniMax

1363

New contributor

$begingroup$
Thank you @MinMax. 200_success also gave the same answer earlier.
$endgroup$
– RebornCodeLover
9 hours ago

$begingroup$
@RebornCodeLover It is different. @200_success uses the islice from itertools. I use built-in slice. But yes, they are similar in other ways.
$endgroup$
– MiniMax
9 hours ago

$begingroup$
True. Yes I like it. Thanks
$endgroup$
– RebornCodeLover
9 hours ago

add a comment |

It can be solved by using the slice object.

From Python documentation:

slice - An object usually containing a portion of a sequence. A slice is created using the subscript notation, with colons between numbers
when several are given, such as in variable_name[1:3:5]. The bracket
(subscript) notation uses slice objects internally.

cols = 4 

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



end = step * cols

for start in range(step):

    print(', '.join(vlist[start:end:step]))

Output

Value1, Value11, Value21, Value31

Value2, Value12, Value22, Value32

Value3, Value13, Value23, Value33

Value4, Value14, Value24, Value34

Value5, Value15, Value25, Value35

Value6, Value16, Value26, Value36

Value7, Value17, Value27, Value37

Value8, Value18, Value28, Value38

Value9, Value19, Value29, Value39

Value10, Value20, Value30, Value40

answered 9 hours ago

MiniMax

1363

New contributor

It can be solved by using the slice object.

From Python documentation:

slice - An object usually containing a portion of a sequence. A slice is created using the subscript notation, with colons between numbers
when several are given, such as in variable_name[1:3:5]. The bracket
(subscript) notation uses slice objects internally.

cols = 4 

step = 10

vlist = ["Value" + str(i+1) for i in range(100)]



end = step * cols

for start in range(step):

    print(', '.join(vlist[start:end:step]))

Output

Value1, Value11, Value21, Value31

Value2, Value12, Value22, Value32

Value3, Value13, Value23, Value33

Value4, Value14, Value24, Value34

Value5, Value15, Value25, Value35

Value6, Value16, Value26, Value36

Value7, Value17, Value27, Value37

Value8, Value18, Value28, Value38

Value9, Value19, Value29, Value39

Value10, Value20, Value30, Value40

answered 9 hours ago

MiniMax

1363

New contributor

answered 9 hours ago

MiniMax

1363

New contributor

answered 9 hours ago

MiniMax

1363

answered 9 hours ago

MiniMax

1363

New contributor

MiniMax is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$begingroup$
Thank you @MinMax. 200_success also gave the same answer earlier.
$endgroup$
– RebornCodeLover
9 hours ago

$begingroup$
@RebornCodeLover It is different. @200_success uses the islice from itertools. I use built-in slice. But yes, they are similar in other ways.
$endgroup$
– MiniMax
9 hours ago

$begingroup$
True. Yes I like it. Thanks
$endgroup$
– RebornCodeLover
9 hours ago

add a comment |

$begingroup$
Thank you @MinMax. 200_success also gave the same answer earlier.
$endgroup$
– RebornCodeLover
9 hours ago

$begingroup$
@RebornCodeLover It is different. @200_success uses the islice from itertools. I use built-in slice. But yes, they are similar in other ways.
$endgroup$
– MiniMax
9 hours ago

$begingroup$
True. Yes I like it. Thanks
$endgroup$
– RebornCodeLover
9 hours ago

Thank you @MinMax. 200_success also gave the same answer earlier.

– RebornCodeLover
9 hours ago

@RebornCodeLover It is different. @200_success uses the islice from itertools. I use built-in slice. But yes, they are similar in other ways.

– MiniMax
9 hours ago

True. Yes I like it. Thanks

– RebornCodeLover
9 hours ago

add a comment |

RebornCodeLover is a new contributor. Be nice, and check out our Code of Conduct.

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