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
Related
I am working on a small task in which I have to find the distance between two nodes. Each node has X and Y coordinates which can be seen below.
node_number X_coordinate Y_coordinate
0 0 1 0
1 1 1 1
2 2 1 2
3 3 1 3
4 4 0 3
5 5 0 4
6 6 1 4
7 7 2 4
8 8 3 4
9 9 4 4
10 10 4 3
11 11 3 3
12 12 2 3
13 13 2 2
14 14 2 1
15 15 2 0
For the purpose I mentioned above, I wrote below code,
X1_coordinate = df['X_coordinate'].tolist()
Y1_coordinate = df['Y_coordinate'].tolist()
node_number1 = df['node_number'].tolist()
nodal_dist = []
i = 0
for i in range(len(node_number1)):
dist = math.sqrt((X1_coordinate[i+1] - X1_coordinate[i])**2 + (Y1_coordinate[i+1] - Y1_coordinate[i])**2)
nodal_dist.append(dist)
I got the error
list index out of range
Kindly let me know what I am doing wrong and what should I change to get the answer.
Indexing starts at zero, so the last element in the list has an index that is one less than the number of elements in that list. But the len() function gives you the number of elements in the list (in other words, it starts counting at 1), so you want the range of your loop to be len(node_number1) - 1 to avoid an -off-by-one error.
The problems should been in this line
dist = math.sqrt((X1_coordinate[i+1] - X1_coordinate[i])**2 + (Y1_coordinate[i+1] - Y1_coordinate[i])**2)
the X1_coordinate[i+1] and the ] Y1_coordinate[i+1]] go out of range on the last number call.
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
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 have a sheet of numbers, separated by spaces into columns. Each column represents a different category, and within each column, each number represents a different value. For example, column number four represents age, and within the column, the number 5 represents an age of 44-55. Obviously, each row is a different person's record. I'd like to use a Python script to search through the the sheet, and find all columns where the sixth column is number "1." After that, I want to know how many times each number in column one appears where the number in column six is equal to "1." The script should output to the user that "While column six equals '1', the value '1' appears 12 times in column one. The value '2' appears 18 times..." etc. I hope I'm being clear here. I just want it to list the numbers, basically. Anyway, I'm new to Python. I've attached my code below. I think I should be using dictionaries, but I'm just not totally sure how. So far, I haven't really come close to figuring this out. I would really appreciate if someone could walk me through the logic that would be behind such code. Thank you so much!
ldata = open("list.data", "r")
income_dist = {}
for line in ldata:
linelist = line.strip().split(" ")
key_income_dist = linelist[6]
if key_income_dist in income_dist:
income_dist[key_income_dist] = 1 + income_dist[key_income_dist]
else:
income_dist[key_income_dist] = 1
ldata.close()
print value_no_occupations
First, indentation is majorly important in Python and the above is bad: the 5 lines following linelist = line.strip().split(" ") need to be indented to be in the loop like they should be.
Next they should be indented further and this line added before them:
if len(linelist)>6 and linelist[6]=="1":
This line skips over short lines (there are some), and tests for what you said you wanted: "where column six equals "1."" This is column [6] where the first number on the line is referenced as [0] (these are "offsets", not "cardinal", or counting, numbers).
You'll probably want to change key_income_dist = linelist[6] to key_income_dist = linelist[0] or [1] to get what you want. Play around if necessary.
Finally, you should say print income_dist at the end to get a look at your results. If you want fancier output, study up on formatting.
This is actually easier than it seems! The key is collections.Counter
from collections import Counter
ldata = open("list.data")
rows = [tuple(row.split()) for row in ldata if row.split()[5]==1]
# warning this will break if some rows are shorter than 6 columns
first_col = Counter(item[0] for item in rows)
If you want the distribution of every column (not just the first) do:
distribution = {column: Counter(item[column] for item in rows) for column in range(len(rows[0]))}
# warning this will break if all rows are not the same size!
Considering that the data file has ~9000 rows of data, if you don't want to keep the original data, you can combine step 1 & 2 to make the program use less memory and a little faster.
ldata = open("list.data", "r")
# read in all the rows, note that the list values are strings instead of integers
# keep only the rows with 6th column = '1'
only1 = []
for line in ldata:
if line.strip() == '': # ignor blank lines
continue
row = tuple(line.strip().split(" "))
if row[5] == '1':
only1.append(row)
ldata.close()
# tally the statistics
income_dist = {}
for row in only1:
if row[0] in income_dist:
income_dist[row[0]] += 1
else:
income_dist[row[0]] = 1
# print result
print "While column six equals '1',"
for num in sorted(income_dist):
print "the value %s appears %d times in column one." % (num, income_dist[num])
Sample Test Data in list.data:
9 2 1 5 4 5 5 3 3 0 1 1 7 NA
9 1 1 5 5 5 5 3 5 2 1 1 7 1
9 2 1 3 5 1 5 2 3 1 2 3 7 1
1 2 5 1 2 6 5 1 4 2 3 1 7 1
1 2 5 1 2 6 3 1 4 2 3 1 7 1
8 1 1 6 4 8 5 3 2 0 1 1 7 1
1 1 5 2 3 9 4 1 3 1 2 3 7 1
6 1 3 3 4 1 5 1 1 0 2 3 7 1
2 1 1 6 3 8 5 3 3 0 2 3 7 1
4 1 1 7 4 8 4 3 2 0 2 3 7 1
1 1 5 2 4 1 5 1 1 0 2 3 7 1
4 2 2 2 3 2 5 1 2 0 1 1 5 1
8 2 1 3 6 6 2 2 4 2 1 1 7 1
7 2 1 5 3 5 5 3 4 0 2 1 7 1
1 1 5 2 3 9 4 1 3 1 2 3 7 1
6 1 3 3 4 1 5 1 1 0 2 3 7 1
2 1 1 6 3 8 5 3 3 0 2 3 7 1
4 1 1 7 4 8 4 3 2 0 2 3 7 1
1 1 5 2 4 9 5 1 1 0 2 3 7 1
4 2 2 2 3 2 5 1 2 0 1 1 5 1
Following your original program logic, I come up with this version:
ldata = open("list.data", "r")
# read in all the rows, note that the list values are strings instead of integers
linelist = []
for line in ldata:
linelist.append(tuple(line.strip().split(" ")))
ldata.close()
# keep only the rows with 6th column = '1'
only1 = []
for row in linelist:
if row[5] == '1':
only1.append(row)
# tally the statistics
income_dist = {}
for row in only1:
if row[0] in income_dist:
income_dist[row[0]] += 1
else:
income_dist[row[0]] = 1
# print result
print "While column six equals '1',"
for num in sorted(income_dist):
print "the value %s appears %d times in column one." % (num, income_dist[num])
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