I was wondering how I could align every item in one list, to the corresponding index in the second list. Here is my code so far:
letters=['a','ab','abc','abcd','abcde','abcdef','abcdefg','abcdefgh','abcdefghi','abcdefghij']
numbers=[1,2,3,4,5,6,7,8,9,10]
for x in range(len(letters)):
print letters[x]+"----------",numbers[x]
This is the output I get:
a---------- 1
ab---------- 2
abc---------- 3
abcd---------- 4
abcde---------- 5
abcdef---------- 6
abcdefg---------- 7
abcdefgh---------- 8
abcdefghi---------- 9
abcdefghij---------- 10
This is the output I want:
a---------- 1
ab--------- 2
abc-------- 3
abcd------- 4
abcde------ 5
abcdef----- 6
abcdefg---- 7
abcdefgh--- 8
abcdefghi-- 9
abcdefghij- 10
You could use string formatting:
for left, right in zip(letters, numbers):
print '{0:-<12} {1}'.format(left, right)
And the output:
a----------- 1
ab---------- 2
abc--------- 3
abcd-------- 4
abcde------- 5
abcdef------ 6
abcdefg----- 7
abcdefgh---- 8
abcdefghi--- 9
abcdefghij-- 10
Something like this using string.formatting:
def solve(letters,numbers):
it=iter(range( max(numbers) ,0,-1))
for x,y in zip(letters,numbers):
print "{0}{1} {2}".format(x,"-"*next(it),y)
....:
In [38]: solve(letters,numbers)
a---------- 1
ab--------- 2
abc-------- 3
abcd------- 4
abcde------ 5
abcdef----- 6
abcdefg---- 7
abcdefgh--- 8
abcdefghi-- 9
abcdefghij- 10
letters=['a','ab','abc','abcd','abcde','abcdef','abcdefg','abcdefgh','abcdefghi','abcdefghij']
for c,x in enumerate(letters, start=1):
print x+("-"*(10-c))+" %s" % c
a--------- 1
ab-------- 2
abc------- 3
abcd------ 4
abcde----- 5
abcdef---- 6
abcdefg--- 7
abcdefgh-- 8
abcdefghi- 9
abcdefghij 10
letters=['a','ab','abc','abcd','abcde','abcdef','abcdefg','abcdefgh','abcdefghi','abcdefghij']
numbers=[1,2,3,4,5,6,7,8,9,10]
for x in range(len(letters)):
print '{0:11}{1}'.format(letters[x],numbers[x]).replace(' ','-');
a----------1
ab---------2
abc--------3
abcd-------4
abcde------5
abcdef-----6
abcdefg----7
abcdefgh---8
abcdefghi--9
abcdefghij-10
Related
I would like to create the following pattern:
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
Here is my attempt:
def print_numbers(number):
for i in range(number):
print(i * i)
Now clearly this will return the product of the number with itself. I would like to specify this function to print out the number that many times. So something like print(i) * i which is not correct syntax of course. How would I about doing this?
As mentioned by #Barmar you can convert the number into strings and multiply with same number.
#Ex str(4)*3 = 444 #It will print 3times 4.
def print_numbers(number):
for i in range(number+1):
print((str(i)+" ")*i) #Converted number into string and multiply with the same number.
print_numbers(5)
Output:
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
If you wanted a line space b/w each number you can print like: print((str(i)+" ")*i+"\n") This will give.
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
Assume I have the following pandas data frame:
my_class value
0 1 1
1 1 2
2 1 3
3 2 4
4 2 5
5 2 6
6 2 7
7 2 8
8 2 9
9 3 10
10 3 11
11 3 12
I want to identify the indices of "my_class" where the class changes and remove n rows after and before this index. The output of this example (with n=2) should look like:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
My approach:
# where class changes happen
s = df['my_class'].ne(df['my_class'].shift(-1).fillna(df['my_class']))
# mask with `bfill` and `ffill`
df[~(s.where(s).bfill(limit=1).ffill(limit=2).eq(1))]
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
One of possible solutions is to:
Make use of the fact that the index contains consecutive integers.
Find index values where class changes.
For each such index generate a sequence of indices from n-2
to n+1 and concatenate them.
Retrieve rows with indices not in this list.
The code to do it is:
ind = df[df['my_class'].diff().fillna(0, downcast='infer') == 1].index
df[~df.index.isin([item for sublist in
[ range(i-2, i+2) for i in ind ] for item in sublist])]
my_class = np.array([1] * 3 + [2] * 6 + [3] * 3)
cols = np.c_[my_class, np.arange(len(my_class)) + 1]
df = pd.DataFrame(cols, columns=['my_class', 'value'])
df['diff'] = df['my_class'].diff().fillna(0)
idx2drop = []
for i in df[df['diff'] == 1].index:
idx2drop += range(i - 2, i + 2)
print(df.drop(idx_drop)[['my_class', 'value']])
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12
Here's the program:
layout = "{0:>5}"
layout += "{1:>10}"
for i in range(2, 13):
layout += "{"+str(i)+":9>}"
index = []
for i in range(13):
index.append(i)
index = tuple(index)
print(layout.format(*index))
and it prints out like this:
0 123456789101112
but I want it to look something like this(number of spaces might be wrong):
0 1 2 3 4 5 6 7 8 9 10 11 12
What did I do wrong?
":9>}"
should be
":>9}"
This gives:
0 1 2 3 4 5 6 7 8 9 10 11 12
To look exactly like what you ask:
Actually, you're asking for something weird, but here's a more succint way to write what you wrote:
layout = "{0:>5}{1:>5}" + ''.join("{" + str(i) + ":>4}" for i in range(2, 13))
print(layout.format(*range(13)))
Gives:
0 1 2 3 4 5 6 7 8 9 10 11 12
I am stuck trying to print out a table in Python which would look like this (first number stands for amount of numbers, second for amount of columns):
>>> print_table(13,4)
0 1 2 3
4 5 6 7
8 9 10 11
12 13
Does anyone know a way to achieve this?
This is slightly more difficult than it sounds initially.
def numbers(n, r):
print('\n'.join(' '.join(map(str, range(r*i, min(r*(i + 1), n + 1)))) for i in range(n//r + 1)))
numbers(13, 4)
#>>> 0 1 2 3
4 5 6 7
8 9 10 11
12 13
def numbers(a,b):
i=0;
c=0;
while i<=a:
print(i,end="") #prevents printing a new line
c+=1
if c>=b:
print("\n") #prints a new line when the number of columns is reached and then reset the current column number
c=0;
I think it should work
def num2(n=10, r=3):
print('\n'.join(' '.join(tuple(map(str, range(n+1)))[i:i+r]) for i in range(0, n+1, r)))
<<<
0 1 2
3 4 5
6 7 8
9 10
I'm trying to print a nested loops that looks like this:
1 2 3 4
5 6 7 8
9 10 11 12
This is what I have so far:
def main11():
for n in range(1,13)
print(n, end=' ')
however, this prints the numbers in one line: 1 2 3 4 5 6 7 8 9 10 11 12
You can do that using string formatting:
for i in range(1,13):
print '{:2}'.format(i),
if i%4==0: print
[OUTPUT]
1 2 3 4
5 6 7 8
9 10 11 12
Modulus Operator (%)
for n in range(1,13):
print(n, end=' ')
if n%4 == 0:
print
for offset in range(3):
for i in range(1,5):
n = offset*4 + i
print(n, end=' ')
print()
Output:
1 2 3 4
5 6 7 8
9 10 11 12
Or if you want it nicely formatted the way you have in your post:
for offset in range(3):
for i in range(1,5):
n = offset*4 + i
print("% 2s"%n, end=' ')
print()
Output:
1 2 3 4
5 6 7 8
9 10 11 12
Most of the time when you write a for loop, you should check if this is the right implementation.
From the requirements you have, I would write something like this:
NB_NB_INLINE = 4
MAX_NB = 12
start = 1
while start < MAX_NB:
print( ("{: 3d}" * NB_NB_INLINE).format(*tuple( j+start for j in range(NB_NB_INLINE))) )
start += NB_NB_INLINE