for j in range(10):
for i in range(10):
print(j,end=" ")
My results are bunched together and I need to have 10 numbers per line. I cant use a print("0123456789"). I have tried print(j,j,j,j,j,j,j,j,j) and I get the results that I'm looking for but I'm sure this isn't the proper way to write the code.
If print(j,j,j,j,j,j,j,j,j) works then you simply need to add another print() after each iteration:
for j in range(10):
for i in range(10):
print(j,end=" ")
print()
Output:
0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4 4 4
5 5 5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9
Or simply:
for j in range(10):
print(" ".join(str(j) * 10))
0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4 4 4
5 5 5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9
Why are you using a nested for loop when you can use a single for loop:
for i in range(10):
print('{} '.format(i) * 10)
This is similar to Malik Brahimi's solution, except it doesn't put a space after the last digit on each line:
for i in range(10):
print(' '.join([str(i)]*10))
output
0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4 4 4
5 5 5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9
Just for fun, here's another way to do it with a single loop, this time using a format string with numbered fields.
fmt = ('{0} ' * 10)[:-1]
for i in range(10):
print(fmt.format(i))
Related
1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 7 6 5 4 3 2 1
1 2 3 4 5 6 6 5 4 3 2 1
1 2 3 4 5 5 4 3 2 1
1 2 3 4 4 3 2 1
1 2 3 3 2 1
1 2 2 1
1 1
How can I code this in python?
So far this is my code but I can't figure out the spaces.
number = int(input("enter a number to create your triangle: "))
for col in range(number,0,-1):
for row in range(1,col):
print(row, end=" ")
if col<number:
print(" "*(row*2), end="")
for row in range(col-1,0,-1):
print(row, end=" ")
print()
Instead of looping multiple times, you could create a base list containing all numbers as strings, iterate once to print each row and the reverse:
number = 9
base = [str(n) for n in range(1, number+1)]
for i, idx in enumerate(range(len(base)-1, 0, -1)):
if i == 0:
print(' '.join(base + base[::-1]))
base[idx] = ' '
print(' '.join(base + base[::-1]))
Out:
1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 7 6 5 4 3 2 1
1 2 3 4 5 6 6 5 4 3 2 1
1 2 3 4 5 5 4 3 2 1
1 2 3 4 4 3 2 1
1 2 3 3 2 1
1 2 2 1
1 1
As I've pointed in this comment, using all "power" of print() and unpacking you can make it really easy.
By default all arguments passed to print() will be separated with space, so we can just unpack there range from 1 to current index, string of spaces and reverse range from current index to 1:
number = int(input("enter a number to create your triangle: "))
for i, n in enumerate(range(number, 0, -1)):
print(*range(1, n + 1), *(" " * (i * 2)), *range(n, 0, -1))
Output:
1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 7 6 5 4 3 2 1
1 2 3 4 5 6 6 5 4 3 2 1
1 2 3 4 5 5 4 3 2 1
1 2 3 4 4 3 2 1
1 2 3 3 2 1
1 2 2 1
1 1
Important notice: Code above will work properly only for number below 10.
df = pd.DataFrame({'site':[1,1,1,1,1,1,1,1,1,1], 'parm':[8,8,8,8,8,9,9,9,9,9],
'date':[1,2,3,4,5,1,2,3,4,5], 'obs':[1,1,2,3,3,3,5,5,6,6]})
Output
site parm date obs
0 1 8 1 1
1 1 8 2 1
2 1 8 3 2
3 1 8 4 3
4 1 8 5 3
5 1 9 1 3
6 1 9 2 5
7 1 9 3 5
8 1 9 4 6
9 1 9 5 6
I want to count repeating, sequential "obs" values within a "site" and "parm". I have this code which is close:
df['consecutive'] = df.parm.groupby((df.obs != df.obs.shift()).cumsum()).transform('size')
Output
site parm date obs consecutive
0 1 8 1 1 2
1 1 8 2 1 2
2 1 8 3 2 1
3 1 8 4 3 3
4 1 8 5 3 3
5 1 9 1 3 3
6 1 9 2 5 2
7 1 9 3 5 2
8 1 9 4 6 2
9 1 9 5 6 2
It creates the new column with the count. The gap is when the parm changes from 8 to 9 it includes the parm 9 in the parm 8 count. The expected output is:
site parm date obs consecutive
0 1 8 1 1 2
1 1 8 2 1 2
2 1 8 3 2 1
3 1 8 4 3 2
4 1 8 5 3 2
5 1 9 1 3 1
6 1 9 2 5 2
7 1 9 3 5 2
8 1 9 4 6 2
9 1 9 5 6 2
You need to throw site, parm as indicated in the question into groupby:
df['consecutive'] = (df.groupby([df.obs.ne(df.obs.shift()).cumsum(),
'site', 'parm']
)
['obs'].transform('size')
)
Output:
site parm date obs consecutive
0 1 8 1 1 2
1 1 8 2 1 2
2 1 8 3 2 1
3 1 8 4 3 2
4 1 8 5 3 2
5 1 9 1 3 1
6 1 9 2 5 2
7 1 9 3 5 2
8 1 9 4 6 2
9 1 9 5 6 2
Here is my code which prints a particular number pattern. I want my number pattern to be in perfect triangular arrangement like:
a = int(input('Enter number: '))
base = a
while base > 0:
for j in range(1, a + 1):
print(' ' * (2 * j - 2), end = '')
for i in range(1, base + 1):
print(str(i), end = ' ')
print()
base -= 1
The output:
Enter number: 5
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
Enter number: 7
1 2 3 4 5 6 7
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
The program works fine for numbers < 10 but when I input a number > 10 it gives a distorted pattern.
For example:
Enter number: 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2 3 4 5 6 7 8 9 10 11 12 13
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
So is there a way to make the pattern right?
If you want to have the same result for two digit numbers, you have to format your string. Here how it also works for two digit results:
a = int(input('Enter number: '))
base = a
while base > 0:
for j in range(1, a + 1):
print(' ' * (2 * j - 2), end = '')
for i in range(1, base + 1):
print('{0:>2}'.format(str(i)), end = ' ')
print()
base -= 1
Result for 15:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2 3 4 5 6 7 8 9 10 11 12 13
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
Some adjustments and str.rjust will do the trick:
a = base = 15
while base > 0:
for j in range(a):
print(' ' * 3 * j, end='')
for i in range(base):
print(str(i+1).rjust(3), end='')
print()
base -= 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2 3 4 5 6 7 8 9 10 11 12 13
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
You might use rjust method of str, it does:
Return a right-justified string of length width. Padding is done using
the specified fill character (default is a space).
Simple example usage:
numbers = [1, 10, 10, 1000, 10000]
for n in numbers:
print(str(n).rjust(5))
Output:
1
10
10
1000
10000
Note that rjust requires at least one argument: width, if original str is shorther than width leading spaces (or other characters if specified) will be added to get str of length equal to width, otherwise original str will be returned.
Assume I have a dataframe like this, for example:
0 1 2 3 4 5 6 7 8 9
0 8 9 2 1 6 2 6 8 6 3
1 1 1 8 3 1 6 3 6 3 9
2 1 4 3 5 9 3 5 9 2 3
3 4 6 3 8 4 3 1 5 1 1
4 1 8 5 3 9 6 1 7 2 2
5 6 6 7 9 1 8 2 3 2 8
6 8 3 6 9 9 5 8 4 7 7
7 8 3 3 8 7 1 4 9 7 2
8 7 6 1 4 8 1 6 9 6 6
9 3 3 2 4 8 1 8 1 1 8
10 7 7 5 7 1 4 1 8 8 6
11 6 3 2 7 6 5 7 4 8 7
I would like to put rows to certain "blocks" of given length and the flatten them to single rows. So for example, if the block length would be 3, the result here would be:
0 1 2 3 4 5 6 7 8 9 10 ... 19 20 21 22 23 24 25 26 27 28 29
2 8 9 2 1 6 2 6 8 6 3 1 ... 9 1 4 3 5 9 3 5 9 2 3
5 4 6 3 8 4 3 1 5 1 1 1 ... 2 6 6 7 9 1 8 2 3 2 8
8 8 3 6 9 9 5 8 4 7 7 8 ... 2 7 6 1 4 8 1 6 9 6 6
11 3 3 2 4 8 1 8 1 1 8 7 ... 6 6 3 2 7 6 5 7 4 8 7
How to achieve this?
I think need reshape:
n_blocks =3
df = pd.DataFrame(df.values.reshape(-1, n_blocks *df.shape[1]))
print (df)
0 1 2 3 4 5 6 7 8 9 ... 20 21 22 23 24 25 26 27 \
0 8 9 2 1 6 2 6 8 6 3 ... 1 4 3 5 9 3 5 9
1 4 6 3 8 4 3 1 5 1 1 ... 6 6 7 9 1 8 2 3
2 8 3 6 9 9 5 8 4 7 7 ... 7 6 1 4 8 1 6 9
3 3 3 2 4 8 1 8 1 1 8 ... 6 3 2 7 6 5 7 4
28 29
0 2 3
1 2 8
2 6 6
3 8 7
[4 rows x 30 columns]
I found this solution, maybe someone comes up with a better one:
def toBlocks(df, blocklen):
shifted = [df.shift(periods=p) for p in range(blocklen)]
return pd.concat(shifted, axis=1)[blocklen-1:]
This question already has answers here:
Set values on the diagonal of pandas.DataFrame
(8 answers)
Closed 5 years ago.
I have a Pandas Dataframe question. I have a df with index=column. It looks like below.
df:
DNA Cat2
Item A B C D E F F H I J .......
DNA Item
Cat2 A 812 62 174 0 4 46 46 7 2 15
B 62 427 27 0 0 12 61 2 4 11
C 174 27 174 0 0 13 22 5 2 4
D 0 0 0 0 0 0 0 0 0 0
E 4 0 0 0 130 10 57 33 4 5
F 46 12 13 0 10 187 4 5 0 0
......
Another words, df=df.transpose(). All I want to do is find pandas (or numpy for df.values())function to delete index=column values. My ideal output would be below.
df:
DNA Cat2
Item A B C D E F F H I J .......
DNA Item
Cat2 A 0 62 174 0 4 46 46 7 2 15
B 62 0 27 0 0 12 61 2 4 11
C 174 27 0 0 0 13 22 5 2 4
D 0 0 0 0 0 0 0 0 0 0
E 4 0 0 0 0 10 57 33 4 5
F 46 12 13 0 10 0 4 5 0 0
......
Is there a python function that makes this step very fast? I tried for loop with df.iloc[i,i]=0 but since my dataset is ver big, it takes long time to finish. Thanks in advance!
Setup
np.random.seed([3,1415])
i = pd.MultiIndex.from_product(
[['Cat2'], list('ABCDEFGHIJ')],
names=['DNA', 'Item']
)
a = np.random.randint(5, size=(10, 10))
df = pd.DataFrame(a + a.T + 1, i, i)
df
DNA Cat2
Item A B C D E F G H I J
DNA Item
Cat2 A 1 6 6 7 7 7 4 4 8 2
B 6 1 3 6 1 6 6 4 8 5
C 6 3 9 8 9 6 7 8 4 9
D 7 6 8 1 6 9 4 5 4 3
E 7 1 9 6 9 7 3 7 2 6
F 7 6 6 9 7 9 3 4 6 6
G 4 6 7 4 3 3 9 4 5 5
H 4 4 8 5 7 4 4 5 4 5
I 8 8 4 4 2 6 5 4 9 7
J 2 5 9 3 6 6 5 5 7 3
Option 1
Simplest way is to multiply by 1 less the identity
df * (1 - np.eye(len(df), dtype=int))
DNA Cat2
Item A B C D E F G H I J
DNA Item
Cat2 A 0 6 6 7 7 7 4 4 8 2
B 6 0 3 6 1 6 6 4 8 5
C 6 3 0 8 9 6 7 8 4 9
D 7 6 8 0 6 9 4 5 4 3
E 7 1 9 6 0 7 3 7 2 6
F 7 6 6 9 7 0 3 4 6 6
G 4 6 7 4 3 3 0 4 5 5
H 4 4 8 5 7 4 4 0 4 5
I 8 8 4 4 2 6 5 4 0 7
J 2 5 9 3 6 6 5 5 7 0
Option 2
However, we can also use pd.DataFrame.mask with np.eye. Masking is nice because it doesn't have to be numeric and it will still work.
df.mask(np.eye(len(df), dtype=bool), 0)
DNA Cat2
Item A B C D E F G H I J
DNA Item
Cat2 A 0 6 6 7 7 7 4 4 8 2
B 6 0 3 6 1 6 6 4 8 5
C 6 3 0 8 9 6 7 8 4 9
D 7 6 8 0 6 9 4 5 4 3
E 7 1 9 6 0 7 3 7 2 6
F 7 6 6 9 7 0 3 4 6 6
G 4 6 7 4 3 3 0 4 5 5
H 4 4 8 5 7 4 4 0 4 5
I 8 8 4 4 2 6 5 4 0 7
J 2 5 9 3 6 6 5 5 7 0
Option 3
In the event the columns and indices are not identical, OR the are out of order. We can use equality to tell us where to mask.
d = df.iloc[::-1]
d.mask(d.index == d.columns.values[:, None], 0)
DNA Cat2
Item A B C D E F G H I J
DNA Item
Cat2 J 2 5 9 3 6 6 5 5 7 0
I 8 8 4 4 2 6 5 4 0 7
H 4 4 8 5 7 4 4 0 4 5
G 4 6 7 4 3 3 0 4 5 5
F 7 6 6 9 7 0 3 4 6 6
E 7 1 9 6 0 7 3 7 2 6
D 7 6 8 0 6 9 4 5 4 3
C 6 3 0 8 9 6 7 8 4 9
B 6 0 3 6 1 6 6 4 8 5
A 0 6 6 7 7 7 4 4 8 2