Created simple table with for-loop python
Im just trying to create simple table with following this code
n = 0
x = ["list1","list2","list3","list4"]
for y in x:
n+=1
h="{}. {}".format(n,y) + ' ' + "{}. {}".format(n+n,y)
print(h)
And output i got
1. list1 5. list1
2. list2 6. list2
3. list3 7. list3
4. list4 8. list4
What should i do to got output like this
1. list1 3. list3
2. list2 4. list4
Im just beginning and so confused to solved it,thanks for who can answer my issue.
python-3.x for-loop
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
add a comment |
Im just trying to create simple table with following this code
n = 0
x = ["list1","list2","list3","list4"]
for y in x:
n+=1
h="{}. {}".format(n,y) + ' ' + "{}. {}".format(n+n,y)
print(h)
And output i got
1. list1 5. list1
2. list2 6. list2
3. list3 7. list3
4. list4 8. list4
What should i do to got output like this
1. list1 3. list3
2. list2 4. list4
Im just beginning and so confused to solved it,thanks for who can answer my issue.
python-3.x for-loop
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
This is a great problem to ask on here, however, next time try to write the title as a question or stated issue.
– Cory L
Nov 21 '18 at 14:01
Can formatting help here as explained for example at geeksforgeeks.org/python-format-function ?
– petrch
Nov 21 '18 at 15:28
I dont know about this format function can solved my issue,but i can read that maybe can help and thanks for feedback
– katro coplax
Nov 21 '18 at 15:51
add a comment |
Im just trying to create simple table with following this code
n = 0
x = ["list1","list2","list3","list4"]
for y in x:
n+=1
h="{}. {}".format(n,y) + ' ' + "{}. {}".format(n+n,y)
print(h)
And output i got
1. list1 5. list1
2. list2 6. list2
3. list3 7. list3
4. list4 8. list4
What should i do to got output like this
1. list1 3. list3
2. list2 4. list4
Im just beginning and so confused to solved it,thanks for who can answer my issue.
python-3.x for-loop
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
Im just trying to create simple table with following this code
n = 0
x = ["list1","list2","list3","list4"]
for y in x:
n+=1
h="{}. {}".format(n,y) + ' ' + "{}. {}".format(n+n,y)
print(h)
And output i got
1. list1 5. list1
2. list2 6. list2
3. list3 7. list3
4. list4 8. list4
What should i do to got output like this
1. list1 3. list3
2. list2 4. list4
Im just beginning and so confused to solved it,thanks for who can answer my issue.
python-3.x for-loop
python-3.x for-loop
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
edited Nov 21 '18 at 15:24
katro coplax
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
asked Nov 21 '18 at 13:34
katro coplaxkatro coplax
186
186
This is a great problem to ask on here, however, next time try to write the title as a question or stated issue.
– Cory L
Nov 21 '18 at 14:01
Can formatting help here as explained for example at geeksforgeeks.org/python-format-function ?
– petrch
Nov 21 '18 at 15:28
I dont know about this format function can solved my issue,but i can read that maybe can help and thanks for feedback
– katro coplax
Nov 21 '18 at 15:51
add a comment |
This is a great problem to ask on here, however, next time try to write the title as a question or stated issue.
– Cory L
Nov 21 '18 at 14:01
Can formatting help here as explained for example at geeksforgeeks.org/python-format-function ?
– petrch
Nov 21 '18 at 15:28
I dont know about this format function can solved my issue,but i can read that maybe can help and thanks for feedback
– katro coplax
Nov 21 '18 at 15:51
This is a great problem to ask on here, however, next time try to write the title as a question or stated issue.
– Cory L
Nov 21 '18 at 14:01
This is a great problem to ask on here, however, next time try to write the title as a question or stated issue.
– Cory L
Nov 21 '18 at 14:01
Can formatting help here as explained for example at geeksforgeeks.org/python-format-function ?
– petrch
Nov 21 '18 at 15:28
Can formatting help here as explained for example at geeksforgeeks.org/python-format-function ?
– petrch
Nov 21 '18 at 15:28
I dont know about this format function can solved my issue,but i can read that maybe can help and thanks for feedback
– katro coplax
Nov 21 '18 at 15:51
I dont know about this format function can solved my issue,but i can read that maybe can help and thanks for feedback
– katro coplax
Nov 21 '18 at 15:51
add a comment |
2 Answers
2
active
oldest
votes
The below should give you what you are looking for, no matter how long the list is. There may be a more efficient way than this but this way breaks it down into its parts, so essentially you need to find the halfway point in the list and print the respective value on the right.
x = ["list1","list2","list3","list4", "list5",]
if len(x) % 2 == 1 :
half = int((len(x) + 1) / 2)
else:
half = int(len(x) / 2)
for i in range(half):
if half > len(x) / 2:
if i == half - 1:
h = "{}. {}".format (i + 1 , x[i])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1) , x[int (len (x) / 2) + i + 1])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1), x[int (len (x) / 2) + i])
print(h)
I have also taken into account lists that have an odd number of elements like the above example for x, which will give you the following output
1. list1 4. list4
2. list2 5. list5
3. list3
answered Nov 21 '18 at 13:56
Cory LCory L
1599
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
add a comment |
Something like this?
import math
x = ["list1","list2","list3","list4", "list5", "list6", "list7"]
y = math.ceil(len(x)/2)
for i, j in enumerate(x[:math.ceil(len(x)/2)], start=1):
try:
print("{}. {}".format(i, j) + ' '"{}. {}".format(i+math.ceil(len(x)/2), x[y]))
except:
print("{}. {}".format(i, j))
y +=1
Output:
1. list1 5. list5
2. list2 6. list6
3. list3 7. list7
4. list4
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The below should give you what you are looking for, no matter how long the list is. There may be a more efficient way than this but this way breaks it down into its parts, so essentially you need to find the halfway point in the list and print the respective value on the right.
x = ["list1","list2","list3","list4", "list5",]
if len(x) % 2 == 1 :
half = int((len(x) + 1) / 2)
else:
half = int(len(x) / 2)
for i in range(half):
if half > len(x) / 2:
if i == half - 1:
h = "{}. {}".format (i + 1 , x[i])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1) , x[int (len (x) / 2) + i + 1])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1), x[int (len (x) / 2) + i])
print(h)
I have also taken into account lists that have an odd number of elements like the above example for x, which will give you the following output
1. list1 4. list4
2. list2 5. list5
3. list3
answered Nov 21 '18 at 13:56
Cory LCory L
1599
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
add a comment |
The below should give you what you are looking for, no matter how long the list is. There may be a more efficient way than this but this way breaks it down into its parts, so essentially you need to find the halfway point in the list and print the respective value on the right.
x = ["list1","list2","list3","list4", "list5",]
if len(x) % 2 == 1 :
half = int((len(x) + 1) / 2)
else:
half = int(len(x) / 2)
for i in range(half):
if half > len(x) / 2:
if i == half - 1:
h = "{}. {}".format (i + 1 , x[i])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1) , x[int (len (x) / 2) + i + 1])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1), x[int (len (x) / 2) + i])
print(h)
I have also taken into account lists that have an odd number of elements like the above example for x, which will give you the following output
1. list1 4. list4
2. list2 5. list5
3. list3
answered Nov 21 '18 at 13:56
Cory LCory L
1599
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
add a comment |
The below should give you what you are looking for, no matter how long the list is. There may be a more efficient way than this but this way breaks it down into its parts, so essentially you need to find the halfway point in the list and print the respective value on the right.
x = ["list1","list2","list3","list4", "list5",]
if len(x) % 2 == 1 :
half = int((len(x) + 1) / 2)
else:
half = int(len(x) / 2)
for i in range(half):
if half > len(x) / 2:
if i == half - 1:
h = "{}. {}".format (i + 1 , x[i])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1) , x[int (len (x) / 2) + i + 1])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1), x[int (len (x) / 2) + i])
print(h)
I have also taken into account lists that have an odd number of elements like the above example for x, which will give you the following output
1. list1 4. list4
2. list2 5. list5
3. list3
answered Nov 21 '18 at 13:56
Cory LCory L
1599
The below should give you what you are looking for, no matter how long the list is. There may be a more efficient way than this but this way breaks it down into its parts, so essentially you need to find the halfway point in the list and print the respective value on the right.
x = ["list1","list2","list3","list4", "list5",]
if len(x) % 2 == 1 :
half = int((len(x) + 1) / 2)
else:
half = int(len(x) / 2)
for i in range(half):
if half > len(x) / 2:
if i == half - 1:
h = "{}. {}".format (i + 1 , x[i])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1) , x[int (len (x) / 2) + i + 1])
else:
h = "{}. {}".format (i + 1 , x[i]) + ' ' + "{}. {}".format (half + (i + 1), x[int (len (x) / 2) + i])
print(h)
I have also taken into account lists that have an odd number of elements like the above example for x, which will give you the following output
1. list1 4. list4
2. list2 5. list5
3. list3
answered Nov 21 '18 at 13:56
Cory LCory L
1599
answered Nov 21 '18 at 13:56
Cory LCory L
1599
answered Nov 21 '18 at 13:56
Cory LCory L
1599
answered Nov 21 '18 at 13:56
Cory LCory L
1599
1599
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
add a comment |
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
Its so great but so many space for feed,im just do simple task. but it greate for my reference! Thanks sir
– katro coplax
Nov 21 '18 at 15:16
add a comment |
Something like this?
import math
x = ["list1","list2","list3","list4", "list5", "list6", "list7"]
y = math.ceil(len(x)/2)
for i, j in enumerate(x[:math.ceil(len(x)/2)], start=1):
try:
print("{}. {}".format(i, j) + ' '"{}. {}".format(i+math.ceil(len(x)/2), x[y]))
except:
print("{}. {}".format(i, j))
y +=1
Output:
1. list1 5. list5
2. list2 6. list6
3. list3 7. list7
4. list4
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
add a comment |
Something like this?
import math
x = ["list1","list2","list3","list4", "list5", "list6", "list7"]
y = math.ceil(len(x)/2)
for i, j in enumerate(x[:math.ceil(len(x)/2)], start=1):
try:
print("{}. {}".format(i, j) + ' '"{}. {}".format(i+math.ceil(len(x)/2), x[y]))
except:
print("{}. {}".format(i, j))
y +=1
Output:
1. list1 5. list5
2. list2 6. list6
3. list3 7. list7
4. list4
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
add a comment |
Something like this?
import math
x = ["list1","list2","list3","list4", "list5", "list6", "list7"]
y = math.ceil(len(x)/2)
for i, j in enumerate(x[:math.ceil(len(x)/2)], start=1):
try:
print("{}. {}".format(i, j) + ' '"{}. {}".format(i+math.ceil(len(x)/2), x[y]))
except:
print("{}. {}".format(i, j))
y +=1
Output:
1. list1 5. list5
2. list2 6. list6
3. list3 7. list7
4. list4
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
Something like this?
import math
x = ["list1","list2","list3","list4", "list5", "list6", "list7"]
y = math.ceil(len(x)/2)
for i, j in enumerate(x[:math.ceil(len(x)/2)], start=1):
try:
print("{}. {}".format(i, j) + ' '"{}. {}".format(i+math.ceil(len(x)/2), x[y]))
except:
print("{}. {}".format(i, j))
y +=1
Output:
1. list1 5. list5
2. list2 6. list6
3. list3 7. list7
4. list4
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
answered Nov 21 '18 at 14:34
Srce CdeSrce Cde
1,184612
1,184612
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
add a comment |
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
The output is yes,but i think it can be make simple code :)
– katro coplax
Nov 21 '18 at 15:18
add a comment |
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This is a great problem to ask on here, however, next time try to write the title as a question or stated issue.
– Cory L
Nov 21 '18 at 14:01
Can formatting help here as explained for example at geeksforgeeks.org/python-format-function ?
– petrch
Nov 21 '18 at 15:28
I dont know about this format function can solved my issue,but i can read that maybe can help and thanks for feedback
– katro coplax
Nov 21 '18 at 15:51