Trying to let the user input a number, and print a table according to the square of its size. Here's an example.
Size--> 3
0 1 2
3 4 5
6 7 8
Size--> 4
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
Size--> 6
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
Size--> 9
0 1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16 17
18 19 20 21 22 23 24 25 26
27 28 29 30 31 32 33 34 35
36 37 38 39 40 41 42 43 44
45 46 47 48 49 50 51 52 53
54 55 56 57 58 59 60 61 62
63 64 65 66 67 68 69 70 71
72 73 74 75 76 77 78 79 80
Here's is the code that i have tried.
length=int(input('Size--> '))
size=length*length
biglist=[]
for i in range(size):
biglist.append(i)
biglist = [str(i) for i in biglist]
for i in range(0, len(biglist), length):
print(' '.join(biglist[i: i+length]))
but instead here's what i got
Size--> 3
0 1 2
3 4 5
6 7 8
Size--> 4
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
Size--> 6
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
As you can see the rows are not aligned properly like the example.
What's the simplest way of presenting it in a proper alignment? Thx :)
Using .format on string with right aligning.
And strlen is the number of characters required for each number.
length = int(input('Size--> '))
size = length*length
biglist = []
for i in range(size):
biglist.append(i)
biglist = [str(i) for i in biglist]
strlen = len(str(length**2-1))+1
for i in range(0, len(biglist), length):
# print(' '.join(biglist[i: i+length]))
for x in biglist[i: i+length]:
print(f"{x:>{strlen}}", end='')
print()
Related
I would like to find consecutive numbers in column A and Column B in python, Column A should be ascending but Column B is descending. I am attaching an example file.
Input file
nucleotide
Pos_A
Pos_B
Connection_Pos20_Pos102
20
102
Connection_Pos19_Pos102
19
102
Connection_Pos20_Pos101
20
101
Connection_Pos18_Pos102
18
102
Connection_Pos19_Pos101
19
101
Connection_Pos20_Pos100
20
100
Connection_Pos17_Pos102
17
102
Connection_Pos18_Pos101
18
101
Connection_Pos19_Pos100
19
100
Connection_Pos20_Pos99
20
99
Connection_Pos16_Pos102
16
102
Connection_Pos17_Pos101
17
101
Connection_Pos18_Pos100
18
100
Connection_Pos19_Pos99
19
99
Connection_Pos20_Pos98
20
98
Connection_Pos15_Pos102
15
102
Connection_Pos16_Pos101
16
101
Connection_Pos17_Pos100
17
100
Connection_Pos18_Pos99
18
99
Connection_Pos19_Pos98
19
98
Connection_Pos20_Pos97
20
97
Connection_Pos14_Pos102
14
102
Connection_Pos15_Pos101
15
101
Connection_Pos16_Pos100
16
100
Output:
nucleotide
Pos_A
Pos_B
Consecutive ID
Consecutive Number (Size)
Connection_Pos20_Pos102
20
102
101
1
Connection_Pos19_Pos102
19
102
100
2
Connection_Pos20_Pos101
20
101
100
2
Connection_Pos18_Pos102
18
102
99
3
Connection_Pos19_Pos101
19
101
99
3
Connection_Pos20_Pos100
20
100
99
3
Connection_Pos17_Pos102
17
102
98
4
Connection_Pos18_Pos101
18
101
98
4
Connection_Pos19_Pos100
19
100
98
4
Connection_Pos20_Pos99
20
99
98
4
Connection_Pos16_Pos102
16
102
97
5
Connection_Pos17_Pos101
17
101
97
5
Connection_Pos18_Pos100
18
100
97
5
Connection_Pos19_Pos99
19
99
97
5
Connection_Pos20_Pos98
20
98
97
5
Connection_Pos15_Pos102
15
102
96
6
Connection_Pos16_Pos101
16
101
96
6
Connection_Pos17_Pos100
17
100
96
6
Connection_Pos18_Pos99
18
99
96
6
Connection_Pos19_Pos98
19
98
96
6
Connection_Pos20_Pos97
20
97
96
6
Connection_Pos14_Pos102
14
102
95
7
Connection_Pos15_Pos101
15
101
95
7
Connection_Pos16_Pos100
16
100
95
7
Connection_Pos17_Pos99
17
99
95
7
Connection_Pos18_Pos98
18
98
95
7
Connection_Pos19_Pos97
19
97
95
7
Connection_Pos20_Pos96
20
96
95
7
For Consecutive ID, if Pos_B's shifted difference != 1, then we want to subtract 1, so we mark those indexes as -1 with mul(-1) and cumsum them:
df['ID'] = df.Pos_B.shift().sub(df.Pos_B).ne(1).mul(-1).cumsum() + df.Pos_B[0]
For Consecutive Number, if Pos_A's shifted difference != -1, then we want to add 1, so we mark those indexes as 1 and cumsum again:
df['Number'] = df.Pos_A.shift().sub(df.Pos_A).ne(-1).mul(1).cumsum()
Result:
nucleotide Pos_A Pos_B ID Number
0 Connection_Pos20_Pos102 20 102 101 1
1 Connection_Pos19_Pos102 19 102 100 2
2 Connection_Pos20_Pos101 20 101 100 2
3 Connection_Pos18_Pos102 18 102 99 3
4 Connection_Pos19_Pos101 19 101 99 3
5 Connection_Pos20_Pos100 20 100 99 3
6 Connection_Pos17_Pos102 17 102 98 4
7 Connection_Pos18_Pos101 18 101 98 4
8 Connection_Pos19_Pos100 19 100 98 4
9 Connection_Pos20_Pos99 20 99 98 4
10 Connection_Pos16_Pos102 16 102 97 5
11 Connection_Pos17_Pos101 17 101 97 5
12 Connection_Pos18_Pos100 18 100 97 5
13 Connection_Pos19_Pos99 19 99 97 5
14 Connection_Pos20_Pos98 20 98 97 5
15 Connection_Pos15_Pos102 15 102 96 6
16 Connection_Pos16_Pos101 16 101 96 6
17 Connection_Pos17_Pos100 17 100 96 6
18 Connection_Pos18_Pos99 18 99 96 6
19 Connection_Pos19_Pos98 19 98 96 6
20 Connection_Pos20_Pos97 20 97 96 6
21 Connection_Pos14_Pos102 14 102 95 7
22 Connection_Pos15_Pos101 15 101 95 7
23 Connection_Pos16_Pos100 16 100 95 7
Do it one by one then groupby with ngroup
s1 = df.Pos_A.diff().le(0).cumsum()
s2 = df.Pos_B.diff().ge(0).cumsum()
df['out'] = df.groupby([s1,s2]).ngroup()+1
Out[452]:
0 1
1 2
2 2
3 3
4 3
5 3
6 4
7 4
8 4
9 4
10 5
11 5
12 5
13 5
14 5
15 6
16 6
17 6
18 6
19 6
20 6
21 7
22 7
23 7
24 7
25 7
26 7
27 7
dtype: int64
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 2 years ago.
Improve this question
I created a lottery number generator with Python 3.7. It shows however None at the end of each try. Here's my code.
import random
def lotto_gen():
n = 1
while n < 7:
print(random.randint(1, 45), end='\t')
n += 1
return
for numbers in range(100):
print(lotto_gen())
And the result goes like this:
6 12 42 37 13 44 None
36 31 32 41 4 30 None
20 31 38 42 14 19 None
8 18 29 22 34 29 None
26 34 15 1 20 38 None
10 17 28 35 22 38 None
23 34 42 22 4 43 None
25 16 17 36 17 4 None
44 8 20 1 43 43 None
29 32 9 2 8 5 None
16 44 35 17 42 10 None
5 1 39 28 21 40 None
35 25 12 31 23 21 None
13 25 9 10 41 7 None
12 34 14 36 27 5 None
32 30 12 5 41 14 None
23 30 5 30 7 9 None
38 25 6 17 17 20 None
12 1 13 10 30 32 None
15 1 3 23 28 6 None
1 2 24 33 36 31 None
28 13 42 39 9 39 None
41 44 2 9 41 34 None
25 19 30 26 8 44 None
39 36 44 4 22 7 None
7 44 29 38 1 8 None
37 6 44 6 41 11 None
29 29 23 40 23 36 None
25 39 30 40 40 4 None
28 14 33 4 15 34 None
41 35 7 26 30 24 None
10 34 26 45 12 10 None
32 6 45 16 24 18 None
14 7 8 26 32 4 None
22 43 40 3 20 31 None
6 42 38 11 18 20 None
6 40 5 18 25 29 None
37 19 26 19 45 41 None
39 8 17 19 17 22 None
I want to remove that None bool type. Can someone tell me how can I edit my code?
Rakesh has given the correct answer, but I would like to explain why your code isn't working. The problem seems to be that for a particular iteration, your code is only able to generate 6 random numbers. Take note, that you have initialized n=1, inside the function lotto_gen() and as the condition for executing the while loop is n<7, the code inside lotto_gen() executes only 6 times.
Now the reason why you receive None at the end is because you are trying to print the value returned by lotto_gen, but take note, that the return field inside your code's function is empty, hence None is returned by the function and hence that gets printed.
So for correcting the code you only need to initialize n as n=0, and to remove the appearance of the none, don't call the function inside a print statement, and create a list which contains the 7 values of each iteration and return it. So, you'll need to modify the code in this manner:
import random
def lotto_gen():
n = 0
a=[]
while n < 7:
a.append(random.randint(1, 45))
n += 1
return a
for numbers in range(100):
print(lotto_gen())
You can use this approach too and my code will execute faster as well! :P
This is one approach.
Ex:
import random
def lotto_gen():
return "\t".join(str(random.randint(1, 45)) for _ in range(6))
for numbers in range(100):
print(lotto_gen())
I have a list of 1 column and 50 rows.
I want to divide it into 5 segments. And each segment has to become a column of a dataframe. I do not want the NAN to appear (figure2). How can I solve that?
Like this:
df = pd.DataFrame(result_list)
AWA=df[:10]
REM=df[10:20]
S1=df[20:30]
S2=df[30:40]
SWS=df[40:50]
result = pd.concat([AWA, REM, S1, S2, SWS], axis=1)
result
Figure2
You can use numpy's reshape function:
result_list = [i for i in range(50)]
pd.DataFrame(np.reshape(result_list, (10, 5), order='F'))
Out:
0 1 2 3 4
0 0 10 20 30 40
1 1 11 21 31 41
2 2 12 22 32 42
3 3 13 23 33 43
4 4 14 24 34 44
5 5 15 25 35 45
6 6 16 26 36 46
7 7 17 27 37 47
8 8 18 28 38 48
9 9 19 29 39 49
IGNORING columns 1 & 2 (only the rest of the columns); I would like to obtain the occurrence COUNT of UNIQUE EVEN values (ignoring ODD ones) for the following set of data.
I have tried:
awk '{ a[$3, $4, $5, $6, $7]++ } END { for (b in a) { cnt+=1 } {print cnt}}' file
I obtain 76 but I don’t expect this value.
> 0 0
> 1 0 0
> 2 0 2
> 3 0 0 6
> 4 0 0 8
> 5 0 0 10
> 6 0 2 14
> 7 0 2 16
> 8 0 0 6 20
> 9 0 0 8 24
> 10 0 0 8 26
> 11 0 0 10 32
> 12 0 0 10 34
> 13 0 2 14 40
> 14 0 2 16 42
> 15 0 0 8 24 48
> 16 0 0 8 24 50
> 17 0 0 8 26 56
> 18 0 0 10 32 60
> 19 0 0 10 34 64
> 20 0 0 10 34 66
> 21 0 2 14 40 72
> 22 0 0 8 24 48 76
> 23 0 0 8 24 50 82
> 24 0 0 8 26 56 88
> 25 0 0 8 26 56 90
> 26 0 0 10 32 60 96
> 27 0 0 10 32 60 98
> 28 0 0 10 34 64 104
> 29 0 0 10 34 64 106
> 30 0 0 10 34 66 112
> 31 0 0 10 34 66 114
> 0 1
> 1 1 2 5
> 2 1 2
> 3 1 2 12 23 19
> 4 1 2 12 23
> 5 1 2 12
> 6 1 2 12 28
> 7 1 2 12 28 36
> 8 1 2 12 30 47 45
> 9 1 2 12 30 47
> 10 1 2 12 30
> 11 1 2 12 30 52
> 12 1 2 12 28 38
> 13 1 2 12 28 38 62
> 14 1 2 12 28 38 62 68
> 15 1 2 12 30 54 75
> 16 1 2 12 30 54
> 17 1 2 12 30 54 78
> 18 1 2 12 30 54 78 84
> 19 1 2 12 30 54 78 84 92
> 20 1 2 12 28 38 62 70
> 21 1 2 12 28 38 62 70 108
> 22 1 2 12 30 54 80
> 23 1 2 12 30 54 78 86
> 24 1 2 12 30 54 78 86 120
> 25 1 2 12 30 54 78 84 94
> 26 1 2 12 30 54 78 84 94 124
> 27 1 2 12 30 54 78 84 92 102
> 28 1 2 12 30 54 78 84 92 102 128
> 29 1 2 12 28 38 62 70 110
> 30 1 2 12 28 38 62 70 110 130
> 31 1 2 12 28 38 62 70 108 116
> 0 2
> 1 2 2 5
> 2 2 2
> 3 2 2 5 6
> 4 2 2 5 6 18
> 5 2 2 5 6 18 22
> 6 2 2 14
> 7 2 2 16
> 8 2 2 5 6 20
> 9 2 2 5 6 20 44
> 10 2 2 5 6 18 26
> 11 2 2 5 6 18 22 32
> 12 2 2 5 6 18 22 32 58
> 13 2 2 14 40
> 14 2 2 16 42
> 15 2 2 5 6 20 44 50 75
> 16 2 2 5 6 20 44 50
> 17 2 2 5 6 18 26 56
> 18 2 2 5 6 18 22 32 60
> 19 2 2 14 40 72 109 101
> 20 2 2 14 40 72 109
> 21 2 2 14 40 72
> 22 2 2 5 6 20 44 50 80
> 23 2 2 5 6 20 44 50 80 118
> 24 2 2 5 6 20 44 50 80 118 120
> 25 2 2 5 6 20 44 50 80 118 120 122
> 26 2 2 14 40 72 109 101 102 127
> 27 2 2 14 40 72 109 101 102
> 28 2 2 14 40 72 109 101 104
> 29 2 2 14 40 72 116 133 131
> 30 2 2 14 40 72 116 133
> 31 2 2 14 40 72 116
> 0 3
> 1 3 0
> 2 3 0 4
> 3 3 0 6
> 4 3 0 6 18
> 5 3 0 6 18 22
> 6 3 0 4 16 37
> 7 3 0 4 16
> 8 3 0 6 20
> 9 3 0 6 18 26 47
> 10 3 0 6 18 26
> 11 3 0 6 18 22 32
> 12 3 0 6 18 22 32 58
> 13 3 0 4 16 42 69
> 14 3 0 4 16 42
> 15 3 0 6 18 26 47 48
> 16 3 0 6 18 26 47 48 74
> 17 3 0 6 18 26 56
> 18 3 0 6 18 22 32 60
> 19 3 0 6 18 22 32 58 64
> 20 3 0 6 18 22 32 58 66
> 21 3 0 6 18 22 32 58 66 108
> 22 3 0 6 18 26 47 48 76
> 23 3 0 6 18 26 56 86
> 24 3 0 6 18 26 56 88
> 25 3 0 6 18 26 56 90
> 26 3 0 6 18 22 32 60 96
> 27 3 0 6 18 22 32 60 98
> 28 3 0 6 18 22 32 58 64 104
> 29 3 0 6 18 22 32 58 64 106
> 30 3 0 6 18 22 32 58 66 112
> 31 3 0 6 18 22 32 58 66 114
> 0 4
> 1 4 0
> 2 4 2
> 3 4 0 6
> 4 4 0 8
> 5 4 0 10
> 6 4 2 16 37
> 7 4 2 16
> 8 4 0 8 24 45
> 9 4 0 8 24
> 10 4 0 8 26
> 11 4 0 8 26 52
> 12 4 2 16 37 38
> 13 4 2 16 42 69
> 14 4 2 16 42
> 15 4 0 8 24 48
> 16 4 0 8 24 50
> 17 4 0 8 26 56
> 18 4 0 8 26 52 60
> 19 4 2 16 37 38 64
> 20 4 2 16 42 69 70
> 21 4 2 16 42 69 72
> 22 4 0 8 24 48 76
> 23 4 0 8 24 50 82
> 24 4 0 8 26 56 88
> 25 4 0 8 26 52 60 94
> 26 4 0 8 26 52 60 96
> 27 4 0 8 26 52 60 98
> 28 4 2 16 37 38 64 104
> 29 4 2 16 42 69 70 110
> 30 4 2 16 42 69 70 112
> 31 4 2 16 42 69 70 114
You can try this awk command to count unique values ignoring 1st and 2nd column:
awk '{$1=$2=""; !seen[$0]++} END{print length(seen)}' file
130
If you are counting uniques excluding 1st and 2nd columns and ignoring odd numbers then use:
awk '{for (i=3; i<=NF; i++) !($i%2) && !seen[$i]++} END{print length(seen)}' file
63
This question already has answers here:
Calculate Percentile in Excel 2010
(3 answers)
Closed 9 years ago.
I am trying to calculate how many calls came back in 95 percentile of time. Below is my Result Set. I am working with Excel 2010
Milliseconds Number
0 1702
1 15036
2 14262
3 13190
4 9137
5 5635
6 3742
7 2628
8 1899
9 1298
10 963
11 727
12 503
13 415
14 311
15 235
16 204
17 140
18 109
19 83
20 72
21 55
22 52
23 35
24 33
25 25
26 15
27 18
28 14
29 15
30 13
31 19
32 23
33 19
34 21
35 20
36 25
37 26
38 13
39 12
40 10
41 17
42 6
43 7
44 8
45 4
46 7
47 9
48 11
49 12
50 9
51 9
52 9
53 8
54 10
55 10
56 11
57 3
58 7
59 7
60 2
61 5
62 7
63 5
64 5
65 2
66 3
67 2
68 1
70 1
71 2
72 1
73 4
74 1
75 1
76 1
77 3
80 1
81 1
85 1
87 2
93 1
96 1
100 1
107 1
112 1
116 1
125 1
190 1
356 1
450 1
492 1
497 1
554 1
957 1
Just some background what does above information means-
1702 calls came back in 0 milliseconds
15036 calls came back in 1 milliseconds
14262 calls came back in 2 milliseconds
etc etc
So to calculate the 95th percentile from the above data, I am using this formula in excel 2010-
=PERCENTILE.EXC(IF(TRANSPOSE(ROW(INDIRECT("1:"&MAX(H$2:H$96))))<=H$2:H$96,A$2:A$96),0.95)
Can anyone help me whether the way I am doing in Excel 2010 is right or not?
I am getting 95th percentile as 10 by using the above scenario.
Thanks for the help.
that's essentially the same question you asked here and the formula I suggested. As per my last comments in that question - that formula should work OK as long as you use CTRL+SHIFT+ENTER correctly. I get 10 as the answer for this example using that formula.
I think you can verify manually that that is indeed the correct answer. If you have a running total in an adjacent column then you can see where the 95th percentile is reached......