Is it possible to align the spaces and characters of two strings perfectly?
I have two functions, resulting in two strings.
One just adds a " " between a list of digits:
digits = 34567
new_digits = 3 4 5 6 7
The second function takes the string and prints out the index of the string, such that:
digits = 34567
index_of_digits = 1 2 3 4 5
Now the issue that I am having is when the length of the string is greater than 10, the alignment is off:
I am supposed to get something like this:
Please advice.
If your digits are in a list, you can use format to space them uniformly:
L = [3,4,2,5,6,3,6,2,5,1,4,1]
print(''.join([format(n,'3') for n in range(1,len(L)+1)]))
print(''.join([format(n,'3') for n in L]))
Or with f-string formatting (Python 3.6+):
L = [3,4,2,5,6,3,6,2,5,1,4,1]
print(''.join([f'{n+1:3}' for n in range(len(L))]))
print(''.join([f'{n:3}' for n in L]))
Output:
1 2 3 4 5 6 7 8 9 10 11 12
3 4 2 5 6 3 6 2 5 1 4 1
Ref: join, format, range, list comprehensions
Related
Here is my basic code:
import numpy as np
a = numpy.asarray([ [1,2,3], [4,5,6], [7,8,9] ])
I want to print the arrays as follows:
1 4 7
2 5 8
3 6 9
Also, how would I approach the same concept if my a array has 1000 embedded list in it?
Here is a powerful one-line solution without loops:
print('\n'.join(map(lambda line: ' '.join(map(str, line)), a.T)))
a.T transpose the 2D array, the first map encode a line in a string and the second one concatenate the string lines (by using \n between).
This is an alternative version with generators (likely slower):
print('\n'.join(' '.join(str(item) for item in line) for line in a.T))
Yet another solution with one loop (likely even slower):
for line in a.T:
print(' '.join(str(item) for item in line))
Note the last version produce a trailing new line.
You could also make use of str:
print(str(a.T).translate(str.maketrans({'[':'',']':''})))
1 4 7
2 5 8
3 6 9
print(str(a.T).replace('[', '').replace(']',''))
1 4 7
2 5 8
3 6 9
print(str(a.T).translate(str.maketrans({'[':'',']':''})).replace('\n ', '\n'))
1 4 7
2 5 8
3 6 9
I try to write the number of words from the longest line. I was able to write the number of words in each line, but I can't print the maximum number. The max () function do not works. Can anyone help me?
import os
import sys
import numpy as np
with open('demofile.txt') as f:
lines = f.readlines()
for index, value in enumerate(lines):
number_of_words = len(value.split())
print(number_of_words)
demofile.txt
<=4 1 2 3 4 5 6 7 8 9 10 11
<=4 1 2 3 4 5 6 7 8 9
<=4 1 2 3 4 5 6 7 8 9 10 11 sdad adada affg
<=4 1 2 3 4 5 6 7 8 9 10 11
Output:
12
10
15
12
0
0
0
0
0
0
0
0
0
0
0
I also don't understand why it lists the number of words in the next lines where there are no words
If I understood correctly max() function doesn't work because you are searching max of strings so you need to convert them to ints(floats).
lines = [int(x) for x in lines.split(" ")] // converts to ints
maximum = max(lines)// should work now
UPD:
Edited with comment below.
Before:
int(x) for x in lines
Now:
int(x) for x in lines.split(" ")
I have a DataFrame, namely 'traj', as follow:
x y z
0 5 3 4
1 4 2 8
2 1 1 7
3 Some string here
4 This is spam
5 5 7 8
6 9 9 7
... #continues repeatedly a lot with the same strings here in index 3 and 4
79 4 3 3
80 Some string here
I'm defining a function in order to delete useless strings positioned in certain index from the DataFrame. Here is what I'm trying:
def spam(names,df): #names is a list composed, for instance, by "Some" and "This" in 'traj'
return df.drop(index = ([traj[(traj.iloc[:,0] == n)].index for n in names]))
But when I call it it returns the error:
traj_clean = spam(my_list_of_names, traj)
...
KeyError: '[(3,4,...80)] not found in axis'
If I try alone:
traj.drop(index = ([traj[(traj.iloc[:,0] == 'Some')].index for n in names]))
it works.
I solved it in a different way:
df = traj[~traj[:].isin(names)].dropna()
Where names is a list of the terms you wish to delete.
df will contain only rows without these terms
1. The Problem
Given a positive integer n. Print the pattern as shown in sample outputs.
A code has already been provided. You have to understand the logic of the code on your own and try and make changes to the code so that it gives correct output.
1.1 The Specifics
Input: A positive integer n, 1<= n <=9
Output: Pattern as shown in examples below
Sample input:
4
Sample output:
4444444
4333334
4322234
4321234
4322234
4333334
4444444
Sample input:
5
Sample output:
555555555
544444445
543333345
543222345
543212345
543222345
543333345
544444445
555555555
2. My Answer
2.1 My Code
n=int(input())
answer=[[1]]
for i in range(2, n+1):
t=[i]*((2*i)-3)
answer.insert(0, t)
answer.append(t)
for a in answer:
a.insert(0,i)
a.append(i)
print(answer)
outlst = [' '.join([str(c) for c in lst]) for lst in answer]
for a in outlst:
print(a)
2.2 My Output
Input: 4
4 4 4 4 4 4 4 4 4
4 4 3 3 3 3 3 3 3 4 4
4 4 3 3 2 2 2 2 2 3 3 4 4
4 3 2 1 2 3 4
4 4 3 3 2 2 2 2 2 3 3 4 4
4 4 3 3 3 3 3 3 3 4 4
4 4 4 4 4 4 4 4 4
2.3 Desired Output
4444444
4333334
4322234
4321234
4322234
4333334
4444444
Your answer isn't as expected because you add the same object t to the answer list twice:
answer.insert(0, t)
answer.append(t)
More specifically, when you assign t = [i]*(2*i - 3), a new data structure is created, [i, ..., i], and t just points to that data structure. Then you put the pointer t in the answer list twice.
In the for a in answer loop, when you use a.insert(0, i) and a.append(i), you update the data structure a is pointing to. Since you call insert(0, i) and append(i) on both pointers that point to the same data structure, you effectively insert and append i to that data structure twice. That's why you end up with more digits than you need.
Instead, you could run the loop for a in answer for only the top half of the rows in the answer list (and the middle row that has was created without a pair). E.g. for a in answer[:(len(answer)+1)/2].
Other things you could do:
using literals as the arguments instead of reusing the reference, e.g. append([i]*(2*i-3)). The literal expression will create a new data structure every time.
using a copy in one of the calls, e.g. append(t.copy()). The copy method creates a new list object with a "shallow" copy of the data structure.
Also, your output digits are space-separated, because you used a non-empty string in ' '.join(...). You should use the empty string: ''.join(...).
n=5
answer=[[1]]
for i in range(2, n+1):
t=[i]*((2*i)-3)
answer.insert(0, t)
answer.append(t.copy())
for a in answer:
a.insert(0,i)
a.append(i)
answerfinal=[]
for a in answer:
answerfinal.append(str(a).replace(' ','').replace(',','').replace(']','').replace('[',''))
for a in answerfinal:
print(a)
n = int(input())
for i in range(1,n*2):
for j in range(1,n*2):
if i <= j<=n*2-i: print(n-i+1,end='')
elif i>n and i>=j >= n*2 -i : print(i-n+1,end='')
elif j<=n: print(n-j+1,end="")
else: print(j-n+1,end='')
print()
n = int(input())
k = 2*n - 1
for i in range(k):
for j in range(k):
a = i if i<j else j
a = a if a<k-i else k-i-1
a = a if a<k-j else k-j-1
print(n-a, end = '')
print()
Input is a number, e.g. 9 and I want to print decimal, octal, hex and binary value from 1 to 9 like:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
How can I achieve this in python3 using syntax like
dm, oc, hx, bn = len(str(9)), len(bin(9)[2:]), ...
print("{:dm%d} {:oc%s}" % (i, oct(i[2:]))
I mean if number is 999 so I want decimal 10 to be printed like ' 10' and binary equivalent of 999 is 1111100111 so I want 10 like ' 1010'.
You can use str.format() and its mini-language to do the whole thing for you:
for i in range(1, 10):
print("{v} {v:>6o} {v:>6x} {v:>6b}".format(v=i))
Which will print:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
UPDATE: To define field 'widths' in a variable you can use a format-within-format structure:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for i in range(1, 10):
print("{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}".format(v=i, w=w))
Or define a different width variable for each field type if you want them non-uniformly spaced.
Btw. since you've tagged your question with python-3.x, if you're using Python 3.6 or newer, you can use Literal String Interpolation to simplify it even more:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for v in range(1, 10):
print(f"{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}")