Align iterable : error “string cannot switch from manual field specification to automatic field...
-1
This question already has an answer here:
ValueError: cannot switch from manual field specification to automatic field numbering
1 answer
So here is my code :
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
print('{0:<4}! = {1:<4} no.multi. ={} sum of {}! = {} no.multi. = {}'.format(i,f(i),i,i,sum,z))
z = z + (i+1)
a(19)
I need to get this output: 
But instead i get a sort of pyramid like 
I tried to format inside the brackets but i always get the error : "cannot switch from manual field specification to automatic field numbering"
Thank you in advance
python string format string-formatting
asked Nov 21 '18 at 12:37
SamiSami
104
marked as duplicate by Jean-François Fabre♦ python
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Nov 21 '18 at 12:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
-1
This question already has an answer here:
ValueError: cannot switch from manual field specification to automatic field numbering
1 answer
So here is my code :
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
print('{0:<4}! = {1:<4} no.multi. ={} sum of {}! = {} no.multi. = {}'.format(i,f(i),i,i,sum,z))
z = z + (i+1)
a(19)
I need to get this output:
But instead i get a sort of pyramid like
I tried to format inside the brackets but i always get the error : "cannot switch from manual field specification to automatic field numbering"
Thank you in advance
python string format string-formatting
asked Nov 21 '18 at 12:37
SamiSami
104
marked as duplicate by Jean-François Fabre♦ python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.
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Nov 21 '18 at 12:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
-1
-1
-1
This question already has an answer here:
ValueError: cannot switch from manual field specification to automatic field numbering
1 answer
So here is my code :
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
print('{0:<4}! = {1:<4} no.multi. ={} sum of {}! = {} no.multi. = {}'.format(i,f(i),i,i,sum,z))
z = z + (i+1)
a(19)
I need to get this output:
But instead i get a sort of pyramid like
I tried to format inside the brackets but i always get the error : "cannot switch from manual field specification to automatic field numbering"
Thank you in advance
python string format string-formatting
asked Nov 21 '18 at 12:37
SamiSami
104
This question already has an answer here:
ValueError: cannot switch from manual field specification to automatic field numbering
1 answer
So here is my code :
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
print('{0:<4}! = {1:<4} no.multi. ={} sum of {}! = {} no.multi. = {}'.format(i,f(i),i,i,sum,z))
z = z + (i+1)
a(19)
I need to get this output:
But instead i get a sort of pyramid like
I tried to format inside the brackets but i always get the error : "cannot switch from manual field specification to automatic field numbering"
Thank you in advance
This question already has an answer here:
ValueError: cannot switch from manual field specification to automatic field numbering
1 answer
python string format string-formatting
python string format string-formatting
asked Nov 21 '18 at 12:37
SamiSami
104
asked Nov 21 '18 at 12:37
SamiSami
104
asked Nov 21 '18 at 12:37
SamiSami
104
asked Nov 21 '18 at 12:37
SamiSami
104
asked Nov 21 '18 at 12:37
SamiSami
104
104
marked as duplicate by Jean-François Fabre♦ python
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Nov 21 '18 at 12:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jean-François Fabre♦ python
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Nov 21 '18 at 12:41
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1 Answer
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I declared variable strong_i which is string of i with !, to take into account ! when choosing amount of spaces when printing i with !.
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
strong_i = str(i) + '!'
print('{0:<4}= {1:<20} no.multi. = {2:<2} sum of {3:<3} = {4:<20} no.multi. = {5}'.format(strong_i,f(i),i,strong_i,sum,z))
z = z + (i+1)
a(19)
Output:
0! = 1 no.multi. = 0 sum of 0! = 1 no.multi. = 0
1! = 1 no.multi. = 1 sum of 1! = 2 no.multi. = 1
2! = 2 no.multi. = 2 sum of 2! = 4 no.multi. = 3
3! = 6 no.multi. = 3 sum of 3! = 10 no.multi. = 6
4! = 24 no.multi. = 4 sum of 4! = 34 no.multi. = 10
5! = 120 no.multi. = 5 sum of 5! = 154 no.multi. = 15
6! = 720 no.multi. = 6 sum of 6! = 874 no.multi. = 21
7! = 5040 no.multi. = 7 sum of 7! = 5914 no.multi. = 28
8! = 40320 no.multi. = 8 sum of 8! = 46234 no.multi. = 36
9! = 362880 no.multi. = 9 sum of 9! = 409114 no.multi. = 45
10! = 3628800 no.multi. = 10 sum of 10! = 4037914 no.multi. = 55
11! = 39916800 no.multi. = 11 sum of 11! = 43954714 no.multi. = 66
12! = 479001600 no.multi. = 12 sum of 12! = 522956314 no.multi. = 78
13! = 6227020800 no.multi. = 13 sum of 13! = 6749977114 no.multi. = 91
14! = 87178291200 no.multi. = 14 sum of 14! = 93928268314 no.multi. = 105
15! = 1307674368000 no.multi. = 15 sum of 15! = 1401602636314 no.multi. = 120
16! = 20922789888000 no.multi. = 16 sum of 16! = 22324392524314 no.multi. = 136
17! = 355687428096000 no.multi. = 17 sum of 17! = 378011820620314 no.multi. = 153
18! = 6402373705728000 no.multi. = 18 sum of 18! = 6780385526348314 no.multi. = 171
19! = 121645100408832000 no.multi. = 19 sum of 19! = 128425485935180314 no.multi. = 190
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Here's simple exampleprint('{0:<5}!.'.format(10))that gives us10 !.because spaces are placed between our number and '!', but if we would do it this wayprint('{0:<5}.'.format('10!'))we'll get10! .
– Filip Młynarski
Nov 21 '18 at 13:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
I declared variable strong_i which is string of i with !, to take into account ! when choosing amount of spaces when printing i with !.
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
strong_i = str(i) + '!'
print('{0:<4}= {1:<20} no.multi. = {2:<2} sum of {3:<3} = {4:<20} no.multi. = {5}'.format(strong_i,f(i),i,strong_i,sum,z))
z = z + (i+1)
a(19)
Output:
0! = 1 no.multi. = 0 sum of 0! = 1 no.multi. = 0
1! = 1 no.multi. = 1 sum of 1! = 2 no.multi. = 1
2! = 2 no.multi. = 2 sum of 2! = 4 no.multi. = 3
3! = 6 no.multi. = 3 sum of 3! = 10 no.multi. = 6
4! = 24 no.multi. = 4 sum of 4! = 34 no.multi. = 10
5! = 120 no.multi. = 5 sum of 5! = 154 no.multi. = 15
6! = 720 no.multi. = 6 sum of 6! = 874 no.multi. = 21
7! = 5040 no.multi. = 7 sum of 7! = 5914 no.multi. = 28
8! = 40320 no.multi. = 8 sum of 8! = 46234 no.multi. = 36
9! = 362880 no.multi. = 9 sum of 9! = 409114 no.multi. = 45
10! = 3628800 no.multi. = 10 sum of 10! = 4037914 no.multi. = 55
11! = 39916800 no.multi. = 11 sum of 11! = 43954714 no.multi. = 66
12! = 479001600 no.multi. = 12 sum of 12! = 522956314 no.multi. = 78
13! = 6227020800 no.multi. = 13 sum of 13! = 6749977114 no.multi. = 91
14! = 87178291200 no.multi. = 14 sum of 14! = 93928268314 no.multi. = 105
15! = 1307674368000 no.multi. = 15 sum of 15! = 1401602636314 no.multi. = 120
16! = 20922789888000 no.multi. = 16 sum of 16! = 22324392524314 no.multi. = 136
17! = 355687428096000 no.multi. = 17 sum of 17! = 378011820620314 no.multi. = 153
18! = 6402373705728000 no.multi. = 18 sum of 18! = 6780385526348314 no.multi. = 171
19! = 121645100408832000 no.multi. = 19 sum of 19! = 128425485935180314 no.multi. = 190
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Here's simple exampleprint('{0:<5}!.'.format(10))that gives us10 !.because spaces are placed between our number and '!', but if we would do it this wayprint('{0:<5}.'.format('10!'))we'll get10! .
– Filip Młynarski
Nov 21 '18 at 13:07
add a comment |
0
I declared variable strong_i which is string of i with !, to take into account ! when choosing amount of spaces when printing i with !.
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
strong_i = str(i) + '!'
print('{0:<4}= {1:<20} no.multi. = {2:<2} sum of {3:<3} = {4:<20} no.multi. = {5}'.format(strong_i,f(i),i,strong_i,sum,z))
z = z + (i+1)
a(19)
Output:
0! = 1 no.multi. = 0 sum of 0! = 1 no.multi. = 0
1! = 1 no.multi. = 1 sum of 1! = 2 no.multi. = 1
2! = 2 no.multi. = 2 sum of 2! = 4 no.multi. = 3
3! = 6 no.multi. = 3 sum of 3! = 10 no.multi. = 6
4! = 24 no.multi. = 4 sum of 4! = 34 no.multi. = 10
5! = 120 no.multi. = 5 sum of 5! = 154 no.multi. = 15
6! = 720 no.multi. = 6 sum of 6! = 874 no.multi. = 21
7! = 5040 no.multi. = 7 sum of 7! = 5914 no.multi. = 28
8! = 40320 no.multi. = 8 sum of 8! = 46234 no.multi. = 36
9! = 362880 no.multi. = 9 sum of 9! = 409114 no.multi. = 45
10! = 3628800 no.multi. = 10 sum of 10! = 4037914 no.multi. = 55
11! = 39916800 no.multi. = 11 sum of 11! = 43954714 no.multi. = 66
12! = 479001600 no.multi. = 12 sum of 12! = 522956314 no.multi. = 78
13! = 6227020800 no.multi. = 13 sum of 13! = 6749977114 no.multi. = 91
14! = 87178291200 no.multi. = 14 sum of 14! = 93928268314 no.multi. = 105
15! = 1307674368000 no.multi. = 15 sum of 15! = 1401602636314 no.multi. = 120
16! = 20922789888000 no.multi. = 16 sum of 16! = 22324392524314 no.multi. = 136
17! = 355687428096000 no.multi. = 17 sum of 17! = 378011820620314 no.multi. = 153
18! = 6402373705728000 no.multi. = 18 sum of 18! = 6780385526348314 no.multi. = 171
19! = 121645100408832000 no.multi. = 19 sum of 19! = 128425485935180314 no.multi. = 190
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Here's simple exampleprint('{0:<5}!.'.format(10))that gives us10 !.because spaces are placed between our number and '!', but if we would do it this wayprint('{0:<5}.'.format('10!'))we'll get10! .
– Filip Młynarski
Nov 21 '18 at 13:07
add a comment |
0
0
0
I declared variable strong_i which is string of i with !, to take into account ! when choosing amount of spaces when printing i with !.
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
strong_i = str(i) + '!'
print('{0:<4}= {1:<20} no.multi. = {2:<2} sum of {3:<3} = {4:<20} no.multi. = {5}'.format(strong_i,f(i),i,strong_i,sum,z))
z = z + (i+1)
a(19)
Output:
0! = 1 no.multi. = 0 sum of 0! = 1 no.multi. = 0
1! = 1 no.multi. = 1 sum of 1! = 2 no.multi. = 1
2! = 2 no.multi. = 2 sum of 2! = 4 no.multi. = 3
3! = 6 no.multi. = 3 sum of 3! = 10 no.multi. = 6
4! = 24 no.multi. = 4 sum of 4! = 34 no.multi. = 10
5! = 120 no.multi. = 5 sum of 5! = 154 no.multi. = 15
6! = 720 no.multi. = 6 sum of 6! = 874 no.multi. = 21
7! = 5040 no.multi. = 7 sum of 7! = 5914 no.multi. = 28
8! = 40320 no.multi. = 8 sum of 8! = 46234 no.multi. = 36
9! = 362880 no.multi. = 9 sum of 9! = 409114 no.multi. = 45
10! = 3628800 no.multi. = 10 sum of 10! = 4037914 no.multi. = 55
11! = 39916800 no.multi. = 11 sum of 11! = 43954714 no.multi. = 66
12! = 479001600 no.multi. = 12 sum of 12! = 522956314 no.multi. = 78
13! = 6227020800 no.multi. = 13 sum of 13! = 6749977114 no.multi. = 91
14! = 87178291200 no.multi. = 14 sum of 14! = 93928268314 no.multi. = 105
15! = 1307674368000 no.multi. = 15 sum of 15! = 1401602636314 no.multi. = 120
16! = 20922789888000 no.multi. = 16 sum of 16! = 22324392524314 no.multi. = 136
17! = 355687428096000 no.multi. = 17 sum of 17! = 378011820620314 no.multi. = 153
18! = 6402373705728000 no.multi. = 18 sum of 18! = 6780385526348314 no.multi. = 171
19! = 121645100408832000 no.multi. = 19 sum of 19! = 128425485935180314 no.multi. = 190
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
I declared variable strong_i which is string of i with !, to take into account ! when choosing amount of spaces when printing i with !.
def f(n):
if n == 0:
return 1
else:
result = 1
for i in range(1,n+1):
result = result * i
return result
def a(n):
sum = 0
z = 0
for i in range(n+1):
sum += f(i)
strong_i = str(i) + '!'
print('{0:<4}= {1:<20} no.multi. = {2:<2} sum of {3:<3} = {4:<20} no.multi. = {5}'.format(strong_i,f(i),i,strong_i,sum,z))
z = z + (i+1)
a(19)
Output:
0! = 1 no.multi. = 0 sum of 0! = 1 no.multi. = 0
1! = 1 no.multi. = 1 sum of 1! = 2 no.multi. = 1
2! = 2 no.multi. = 2 sum of 2! = 4 no.multi. = 3
3! = 6 no.multi. = 3 sum of 3! = 10 no.multi. = 6
4! = 24 no.multi. = 4 sum of 4! = 34 no.multi. = 10
5! = 120 no.multi. = 5 sum of 5! = 154 no.multi. = 15
6! = 720 no.multi. = 6 sum of 6! = 874 no.multi. = 21
7! = 5040 no.multi. = 7 sum of 7! = 5914 no.multi. = 28
8! = 40320 no.multi. = 8 sum of 8! = 46234 no.multi. = 36
9! = 362880 no.multi. = 9 sum of 9! = 409114 no.multi. = 45
10! = 3628800 no.multi. = 10 sum of 10! = 4037914 no.multi. = 55
11! = 39916800 no.multi. = 11 sum of 11! = 43954714 no.multi. = 66
12! = 479001600 no.multi. = 12 sum of 12! = 522956314 no.multi. = 78
13! = 6227020800 no.multi. = 13 sum of 13! = 6749977114 no.multi. = 91
14! = 87178291200 no.multi. = 14 sum of 14! = 93928268314 no.multi. = 105
15! = 1307674368000 no.multi. = 15 sum of 15! = 1401602636314 no.multi. = 120
16! = 20922789888000 no.multi. = 16 sum of 16! = 22324392524314 no.multi. = 136
17! = 355687428096000 no.multi. = 17 sum of 17! = 378011820620314 no.multi. = 153
18! = 6402373705728000 no.multi. = 18 sum of 18! = 6780385526348314 no.multi. = 171
19! = 121645100408832000 no.multi. = 19 sum of 19! = 128425485935180314 no.multi. = 190
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
answered Nov 21 '18 at 12:45
Filip MłynarskiFilip Młynarski
1,7811414
1,7811414
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Here's simple exampleprint('{0:<5}!.'.format(10))that gives us10 !.because spaces are placed between our number and '!', but if we would do it this wayprint('{0:<5}.'.format('10!'))we'll get10! .
– Filip Młynarski
Nov 21 '18 at 13:07
add a comment |
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Here's simple exampleprint('{0:<5}!.'.format(10))that gives us10 !.because spaces are placed between our number and '!', but if we would do it this wayprint('{0:<5}.'.format('10!'))we'll get10! .
– Filip Młynarski
Nov 21 '18 at 13:07
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Thank you very much, but i still don't understand how the ' ! ' affects the output, in such a way that it needs to be isolated
– Sami
Nov 21 '18 at 13:04
Here's simple example
print('{0:<5}!.'.format(10)) that gives us 10 !. because spaces are placed between our number and '!', but if we would do it this way print('{0:<5}.'.format('10!')) we'll get 10! .– Filip Młynarski
Nov 21 '18 at 13:07
Here's simple example
print('{0:<5}!.'.format(10)) that gives us 10 !. because spaces are placed between our number and '!', but if we would do it this way print('{0:<5}.'.format('10!')) we'll get 10! .– Filip Młynarski
Nov 21 '18 at 13:07
add a comment |
