i am having trouble with aligning outcome values.
Alist = ["1,25,999",
"123.4,56.7890,13.571",
"1,23.45,6,7.8"]
c = 0
while c < len(Alist):
r = 0
tokens = Alist[c].split(',')
while r < len(Alist[c].split(',')):
if '.' in tokens[r]:
print "%7.2f" %float(tokens[r]), " ",
else :
print "%3d" %float(tokens[r]), " ",
r += 1
print
c += 1
I want to print such as
1 25 999
123.40 56.79 13.57
1 23.45 6. 7.80
but somehow it is printing
1
25
999
123.40
56.79
13.57
1
23.45
6
7.8
and i cannot figure out what is wrong with my coding.
after the r+1, you have a lone print statement. it is at the wrong indention level - move it to the left by 4 spaces (or one tab) and it should work fine.
The print statement should'nt in the 2nd while loop. just:
Alist = ["1,25,999",
"123.4,56.7890,13.571",
"1,23.45,6,7.8"]
c = 0
while c < len(Alist):
r = 0
tokens = Alist[c].split(',')
while r < len(Alist[c].split(',')):
if '.' in tokens[r]:
print "%7.2f" %float(tokens[r]), " ",
else :
print "%3d" %float(tokens[r]), " ",
r += 1
print
c += 1
In [59]: %paste
myList = ["1,25,999",
"123.4,56.7890,13.571",
"1,23.45,6,7.8"]
rows = [r.split(',') for r in myList]
widths = {i:max(len(c) for c in col) for i,col in enumerate(itertools.izip_longest(*rows, fillvalue=""))}
for row in rows:
for i,val in enumerate(row):
print " "*((widths[i] - len(val))/2), val, " "*((widths[i] - len(val))/2) if not (widths[i]-len(val))%2 else " "*((widths[i] - len(val)+1)/2),
print
## -- End pasted text --
1 25 999
123.4 56.7890 13.571
1 23.45 6 7.8
Related
I have a string with characters repeated. My Job is to find starting Index and ending index of each unique characters in that string. Below is my code.
import re
x = "aaabbbbcc"
xs = set(x)
for item in xs:
mo = re.search(item,x)
flag = item
m = mo.start()
n = mo.end()
print(flag,m,n)
Output :
a 0 1
b 3 4
c 7 8
Here the end index of the characters are not correct. I understand why it's happening but how can I pass the character to be matched dynamically to the regex search function. For instance if I hardcode the character in the search function it provides the desired output
x = 'aabbbbccc'
xs = set(x)
mo = re.search("[b]+",x)
flag = item
m = mo.start()
n = mo.end()
print(flag,m,n)
output:
b 2 5
The above function is providing correct result but here I can't pass the characters to be matched dynamically.
It will be really a help if someone can let me know how to achieve this any hint will also do. Thanks in advance
String literal formatting to the rescue:
import re
x = "aaabbbbcc"
xs = set(x)
for item in xs:
# for patterns better use raw strings - and format the letter into it
mo = re.search(fr"{item}+",x) # fr and rf work both :) its a raw formatted literal
flag = item
m = mo.start()
n = mo.end()
print(flag,m,n) # fix upper limit by n-1
Output:
a 0 3 # you do see that the upper limit is off by 1?
b 3 7 # see above for fix
c 7 9
Your pattern does not need the [] around the letter - you are matching just one anyhow.
Without regex1:
x = "aaabbbbcc"
last_ch = x[0]
start_idx = 0
# process the remainder
for idx,ch in enumerate(x[1:],1):
if last_ch == ch:
continue
else:
print(last_ch,start_idx, idx-1)
last_ch = ch
start_idx = idx
print(ch,start_idx,idx)
output:
a 0 2 # not off by 1
b 3 6
c 7 8
1RegEx: And now you have 2 problems...
Looking at the output, I'm guessing that another option would be,
import re
x = "aaabbbbcc"
xs = re.findall(r"((.)\2*)", x)
start = 0
output = ''
for item in xs:
end = start + len(item[0])
output += (f"{item[1]} {start} {end}\n")
start = end
print(output)
Output
a 0 3
b 3 7
c 7 9
I think it'll be in the Order of N, you can likely benchmark it though, if you like.
import re, time
timer_on = time.time()
for i in range(10000000):
x = "aabbbbccc"
xs = re.findall(r"((.)\2*)", x)
start = 0
output = ''
for item in xs:
end = start + len(item[0])
output += (f"{item[1]} {start} {end}\n")
start = end
timer_off = time.time()
timer_total = timer_off - timer_on
print(timer_total)
So i have for example t=[1.0, 1.0, 1.6, 1.125, 1.5]
I want to print the indices for the elements that have the minimum value min(t) but i want them to start from 1
So for this example i want to print 1 2
It's working when i do this:
for j in range(len(t)):
if t[j]==min(t):
print j+1,`
Output :
1 2
But it's not working with this:
for j in t:
if j==min(t):
a=t.index(j)
print a+1,`
Output :
1 1
Why is that?
Your error is here
a = t.index(j)
t.index(j) is going to give you the first index of that value.
You can make this:
m = min(t)
cont = 0
for j in t:
if j == m:
print cont+1,
cont += 1
Or
for idx, val in enumerate(t):
if val == m:
print idx + 1,
Suppose I have the following local macro:
loc a = 12.000923
I would like to get the decimal position of the first non-zero decimal (4 in this example).
There are many ways to achieve this. One is to treat a as a string and to find the position of .:
loc a = 12.000923
loc b = strpos(string(`a'), ".")
di "`b'"
From here one could further loop through the decimals and count since I get the first non-zero element. Of course this doesn't seem to be a very elegant approach.
Can you suggest a better way to deal with this? Regular expressions perhaps?
Well, I don't know Stata, but according to the documentation, \.(0+)? is suported and it shouldn't be hard to convert this 2 lines JavaScript function in Stata.
It returns the position of the first nonzero decimal or -1 if there is no decimal.
function getNonZeroDecimalPosition(v) {
var v2 = v.replace(/\.(0+)?/, "")
return v2.length !== v.length ? v.length - v2.length : -1
}
Explanation
We remove from input string a dot followed by optional consecutive zeros.
The difference between the lengths of original input string and this new string gives the position of the first nonzero decimal
Demo
Sample Snippet
function getNonZeroDecimalPosition(v) {
var v2 = v.replace(/\.(0+)?/, "")
return v2.length !== v.length ? v.length - v2.length : -1
}
var samples = [
"loc a = 12.00012",
"loc b = 12",
"loc c = 12.012",
"loc d = 1.000012",
"loc e = -10.00012",
"loc f = -10.05012",
"loc g = 0.0012"
]
samples.forEach(function(sample) {
console.log(getNonZeroDecimalPosition(sample))
})
You can do this in mata in one line and without using regular expressions:
foreach x in 124.000923 65.020923 1.000022030 0.0090843 .00000425 {
mata: selectindex(tokens(tokens(st_local("x"), ".")[selectindex(tokens(st_local("x"), ".") :== ".") + 1], "0") :!= "0")[1]
}
4
2
5
3
6
Below, you can see the steps in detail:
. local x = 124.000823
. mata:
: /* Step 1: break Stata's local macro x in tokens using . as a parsing char */
: a = tokens(st_local("x"), ".")
: a
1 2 3
+----------------------------+
1 | 124 . 000823 |
+----------------------------+
: /* Step 2: tokenize the string in a[1,3] using 0 as a parsing char */
: b = tokens(a[3], "0")
: b
1 2 3 4
+-------------------------+
1 | 0 0 0 823 |
+-------------------------+
: /* Step 3: find which values are different from zero */
: c = b :!= "0"
: c
1 2 3 4
+-----------------+
1 | 0 0 0 1 |
+-----------------+
: /* Step 4: find the first index position where this is true */
: d = selectindex(c :!= 0)[1]
: d
4
: end
You can also find the position of the string of interest in Step 2 using the
same logic.
This is the index value after the one for .:
. mata:
: k = selectindex(a :== ".") + 1
: k
3
: end
In which case, Step 2 becomes:
. mata:
:
: b = tokens(a[k], "0")
: b
1 2 3 4
+-------------------------+
1 | 0 0 0 823 |
+-------------------------+
: end
For unexpected cases without decimal:
foreach x in 124.000923 65.020923 1.000022030 12 0.0090843 .00000425 {
if strmatch("`x'", "*.*") mata: selectindex(tokens(tokens(st_local("x"), ".")[selectindex(tokens(st_local("x"), ".") :== ".") + 1], "0") :!= "0")[1]
else display " 0"
}
4
2
5
0
3
6
A straighforward answer uses regular expressions and commands to work with strings.
One can select all decimals, find the first non 0 decimal, and finally find its position:
loc v = "123.000923"
loc v2 = regexr("`v'", "^[0-9]*[/.]", "") // 000923
loc v3 = regexr("`v'", "^[0-9]*[/.][0]*", "") // 923
loc first = substr("`v3'", 1, 1) // 9
loc first_pos = strpos("`v2'", "`first'") // 4: position of 9 in 000923
di "`v2'"
di "`v3'"
di "`first'"
di "`first_pos'"
Which in one step is equivalent to:
loc first_pos2 = strpos(regexr("`v'", "^[0-9]*[/.]", ""), substr(regexr("`v'", "^[0-9]*[/.][0]*", ""), 1, 1))
di "`first_pos2'"
An alternative suggested in another answer is to compare the lenght of the decimals block cleaned from the 0s with that not cleaned.
In one step this is:
loc first_pos3 = strlen(regexr("`v'", "^[0-9]*[/.]", "")) - strlen(regexr("`v'", "^[0-9]*[/.][0]*", "")) + 1
di "`first_pos3'"
Not using regex but log10 instead (which treats a number like a number), this function will:
For numbers >= 1 or numbers <= -1, return with a positive number the number of digits to the left of the decimal.
Or (and more specifically to what you were asking), for numbers between 1 and -1, return with a negative number the number of digits to the right of the decimal where the first non-zero number occurs.
digitsFromDecimal = (n) => {
dFD = Math.log10(Math.abs(n)) | 0;
if (n >= 1 || n <= -1) { dFD++; }
return dFD;
}
var x = [118.8161330, 11.10501660, 9.254180571, -1.245501523, 1, 0, 0.864931613, 0.097007836, -0.010880074, 0.009066729];
x.forEach(element => {
console.log(`${element}, Digits from Decimal: ${digitsFromDecimal(element)}`);
});
// Output
// 118.816133, Digits from Decimal: 3
// 11.1050166, Digits from Decimal: 2
// 9.254180571, Digits from Decimal: 1
// -1.245501523, Digits from Decimal: 1
// 1, Digits from Decimal: 1
// 0, Digits from Decimal: 0
// 0.864931613, Digits from Decimal: 0
// 0.097007836, Digits from Decimal: -1
// -0.010880074, Digits from Decimal: -1
// 0.009066729, Digits from Decimal: -2
Mata solution of Pearly is very likable, but notice should be paid for "unexpected" cases of "no decimal at all".
Besides, the regular expression is not a too bad choice when it could be made in a memorable 1-line.
loc v = "123.000923"
capture local x = regexm("`v'","(\.0*)")*length(regexs(0))
Below code tests with more values of v.
foreach v in 124.000923 605.20923 1.10022030 0.0090843 .00000425 12 .000125 {
capture local x = regexm("`v'","(\.0*)")*length(regexs(0))
di "`v': The wanted number = `x'"
}
I have been stuck on this question for a couple of days. The whole lab question can be found here.
Lab 6: Loopy Lab Part 3
Print the following for any positive integer n. Use an input statement to allow the user to enter the value for n and then print the properly sized box.
E.g. n = 3
1 3 5 5 3 1
3 5 5 3
5 5
5 5
3 5 5 3
1 3 5 5 3 1
E.g. n = 5
1 3 5 7 9 9 7 5 3 1
3 5 7 9 9 7 5 3
5 7 9 9 7 5
7 9 9 7
9 9
9 9
7 9 9 7
5 7 9 9 7 5
3 5 7 9 9 7 5 3
1 3 5 7 9 9 7 5 3 1
Don't worry about handling the spacing for multi-digit numbers.
This is what I have so far:
from __future__ import print_function
for i in range(5):
for j in range(5-i):
print (j, end=" ")
for k in range(i):
print (" "*(2**2-1), end=" ")
for l in range(5-i):
print (l, end=" ")
print()
for i in range(5):
for j in range(i+1):
print (j, end=" ")
print()
I used the import from future function because I am using Python 2.7.3.1. Also, I'm sorry if this seems to be a "noob" question, but I am a beginner and need help. Thank you for all your help! It is appreciated.
This is kind of an interesting question. Here is my solution, not quite optimized but should be straightforward to understand:
n = 5
import string
numberList = string.letters
# uncomment if you want numbers, i prefer letters
#numberList = [i*2-1 for i in range(1, n+1)]
# upper lower
# V V
for n_row in range(0, n) + list(reversed(range(0, n))):
# left number
for number in numberList[:(n-n_row)]:
print number,
# space
for number in range(n_row):
print ' ', ' ',
# right number
for number in reversed(numberList[:(n-n_row)]):
print number,
print
The double digits really screw up the format so i used letters instead.
a b c d e e d c b a
a b c d d c b a
a b c c b a
a b b a
a a
a a
a b b a
a b c c b a
a b c d d c b a
a b c d e e d c b a
Try the following code:
inputValue = int(input('input interger:'))
base = ' '
for i in range(inputValue):
base = ' '+str(2*(inputValue-1-i)+1)+base+str(2*(inputValue-1-i)+1)+' '
length = len(base.strip())
graph = []
for j in range(inputValue):
line = ' '
for k in range(j):
line = ' '+line+' '
for i in range(inputValue-j):
line = ' '+str(2*(inputValue-1-i)+1)+line+str(2*(inputValue-1-i)+1)+' '
line = line.strip()
length_temp = int((length-len(line))/2)
for m in range(length_temp):
line = ' '+line+' '
graph.append(line)
for j in range(inputValue):
graph.append(graph[inputValue-1-j])
for line in graph:
print(line)
Or this code, depending on what your desired output is:
inputValue = int(input('input interger:'))
base = ' '
for i in range(inputValue):
base = ' '+str(2*(inputValue-1-i)+1)+base+str(2*(inputValue-1-i)+1)+' '
length = len(base.strip())
graph = []
for j in range(inputValue):
line = ' '
for k in range(j):
line = ' '+line+' '
for i in range(inputValue-j):
line = ' '+str(2*(inputValue-1-i)+1)+line+str(2*(inputValue-1-i)+1)+' '
line = line.strip()
length_temp = int((length-len(line)))
for m in range(length_temp):
line = line[:int(len(line)/2)]+' '+line[int(len(line)/2)+1:]
graph.append(line)
for j in range(inputValue):
graph.append(graph[inputValue-1-j])
for line in graph:
print(line)
I have a column where each cell has a string of digits, ?, -, and digits in parentheses/brackets/curly brackets. A good example would be something like the following:
3????0{1012}?121-2[101]--01221111(01)1
How do I separate the string into different cells by characters, where a 'character' in this case refers to any number, ?, -, and value within the parentheses/brackets/curly brackets (including said parentheses/brackets/curly brackets)?
In essence, the string above would turn into the following (spaced apart to denote a separate cell):
3 ? ? ? ? 0 {1012} ? 1 2 1 - 2 [101] - - 0 1 2 2 1 1 1 1 (01) 1
The amount of numbers within the parentheses/brackets/curly brackets vary. There are no letters in any of the strings.
Here you are!
RegEx method:
Sub Test_RegEx()
Dim s, col, m
s = "3????0{1012}?121-2[101]--01221111(01)1"
Set col = CreateObject("Scripting.Dictionary")
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "(?:\d|-|\?|\(\d+\)|\[\d+\]|\{\d+\})"
For Each m In .Execute(s)
col(col.Count) = m
Next
End With
MsgBox Join(col.items) ' 3 ? ? ? ? 0 {1012} ? 1 2 1 - 2 [101] - - 0 1 2 2 1 1 1 1 (01) 1
End Sub
Loop method:
Sub Test_Loop()
Dim s, col, q, t, k, i
s = "3????0{1012}?121-2[101]--01221111(01)1"
Set col = CreateObject("Scripting.Dictionary")
q = "_"
t = True
k = 0
For i = 1 To Len(s)
t = (t Or InStr(1, ")]}", q) > 0) And InStr(1, "([{", q) = 0
q = Mid(s, i, 1)
If t Then k = k + 1
col(k) = col(k) & q
Next
MsgBox Join(col.items) ' 3 ? ? ? ? 0 {1012} ? 1 2 1 - 2 [101] - - 0 1 2 2 1 1 1 1 (01) 1
End Sub
Something else to look at :)
Sub test()
'String to parse through
Dim aStr As String
'final string to print
Dim finalString As String
aStr = "3????0{1012}?121-2[101]--01221111(01)1"
'Loop through string
For i = 1 To Len(aStr)
'The character to look at
char = Mid(aStr, i, 1)
'Check if the character is an opening brace, curly brace, or parenthesis
Dim result As String
Select Case char
Case "["
result = loop_until_end(Mid(aStr, i + 1), "]")
i = i + Len(result)
result = char & result
Case "("
result = loop_until_end(Mid(aStr, i + 1), ")")
i = i + Len(result)
result = char & result
Case "{"
result = loop_until_end(Mid(aStr, i + 1), "}")
i = i + Len(result)
result = char & result
Case Else
result = Mid(aStr, i, 1)
End Select
finalString = finalString & result & " "
Next
Debug.Print (finalString)
End Sub
'Loops through and concatenate to a final string until the end_char is found
'Returns a substring starting from the character after
Function loop_until_end(aStr, end_char)
idx = 1
If (Len(aStr) <= 1) Then
loop_until_end = aStr
Else
char = Mid(aStr, idx, 1)
Do Until (char = end_char)
idx = idx + 1
char = Mid(aStr, idx, 1)
Loop
End If
loop_until_end = Mid(aStr, 1, idx)
End Function
Assuming the data is in column A starting in row 1 and that you want the results start in column B and going right for each row of data in column A, here is alternate method using only worksheet formulas.
In cell B1 use this formula:
=IF(OR(LEFT(A1,1)={"(","[","{"}),LEFT(A1,MIN(FIND({")","]","}"},A1&")]}"))),IFERROR(--LEFT(A1,1),LEFT(A1,1)))
In cell C1 use this formula:
=IF(OR(MID($A1,SUMPRODUCT(LEN($B1:B1))+1,1)={"(","[","{"}),MID($A1,SUMPRODUCT(LEN($B1:B1))+1,MIN(FIND({")","]","}"},$A1&")]}",SUMPRODUCT(LEN($B1:B1))+1))-SUMPRODUCT(LEN($B1:B1))),IFERROR(--MID($A1,SUMPRODUCT(LEN($B1:B1))+1,1),MID($A1,SUMPRODUCT(LEN($B1:B1))+1,1)))
Copy the C1 formula right until it starts giving you blanks (there are no more items left to split out from the string in the A cell). In your example, need to copy it right to column AA. Then you can copy the formulas down for the rest of your Column A data.