I'm trying to create a simple game in python 3 and I'm trying to build in an EXP system, for example, every 50 experience points, your health (Which is already an integer) increases by one. Is there a command for this?
(I'm coding this on repl.it if that matters)
I've never shunned guessing. :)
Let me suppose that you are incrementing a variable called experience_points and that, once for every 50 times you increment that you want to increment a variable called health by one.
experience_points += 1
if experience_points % 50 == 0:
health +=1
This bit of code shows how this might work. Notice how health goes up one for every 50 times that 'experience_points` goes up one.
Welcome to the modulus operator!
>>> experience_points = 0
>>> health = 0
>>> while True:
... # do something in the game
... experience_points += 1
... if experience_points % 50 == 0:
... health += 1
... print (experience_points, health, '<--', end='')
... if experience_points > 160:
... break
...
1 0 <--2 0 <--3 0 <--4 0 <--5 0 <--6 0 <--7 0 <--8 0 <--9 0 <--10 0 <--11 0 <--12 0 <--13 0 <--14 0 <--15 0 <--16 0 <--17 0 <--18 0 <--19 0 <--20 0 <--21 0 <--22 0 <--23 0 <--24 0 <--25 0 <--26 0 <--27 0 <--28 0 <--29 0 <--30 0 <--31 0 <--32 0 <--33 0 <--34 0 <--35 0 <--36 0 <--37 0 <--38 0 <--39 0 <--40 0 <--41 0 <--42 0 <--43 0 <--44 0 <--45 0 <--46 0 <--47 0 <--48 0 <--49 0 <--50 1 <--51 1 <--52 1 <--53 1 <--54 1 <--55 1 <--56 1 <--57 1 <--58 1 <--59 1 <--60 1 <--61 1 <--62 1 <--63 1 <--64 1 <--65 1 <--66 1 <--67 1 <--68 1 <--69 1 <--70 1 <--71 1 <--72 1 <--73 1 <--74 1 <--75 1 <--76 1 <--77 1 <--78 1 <--79 1 <--80 1 <--81 1 <--82 1 <--83 1 <--84 1 <--85 1 <--86 1 <--87 1 <--88 1 <--89 1 <--90 1 <--91 1 <--92 1 <--93 1 <--94 1 <--95 1 <--96 1 <--97 1 <--98 1 <--99 1 <--100 2 <--101 2 <--102 2 <--103 2 <--104 2 <--105 2 <--106 2 <--107 2 <--108 2 <--109 2 <--110 2 <--111 2 <--112 2 <--113 2 <--114 2 <--115 2 <--116 2 <--117 2 <--118 2 <--119 2 <--120 2 <--121 2 <--122 2 <--123 2 <--124 2 <--125 2 <--126 2 <--127 2 <--128 2 <--129 2 <--130 2 <--131 2 <--132 2 <--133 2 <--134 2 <--135 2 <--136 2 <--137 2 <--138 2 <--139 2 <--140 2 <--141 2 <--142 2 <--143 2 <--144 2 <--145 2 <--146 2 <--147 2 <--148 2 <--149 2 <--150 3 <--151 3 <--152 3 <--153 3 <--154 3 <--155 3 <--156 3 <--157 3 <--158 3 <--159 3 <--160 3 <--161 3 <--
Related
I have a df as shown below
df:
Id Jan20 Feb20 Mar20 Apr20 May20 Jun20 Jul20 Aug20 Sep20 Oct20 Nov20 Dec20 Amount
1 20 0 0 12 1 3 1 0 0 2 2 0 100
2 0 0 2 1 0 2 0 0 1 0 0 0 500
3 1 2 1 2 3 1 1 2 2 3 1 1 300
From the above I would like to calculate Activeness value which is the number of non zero columns in the month columns as given below.
'Jan20', 'Feb20', 'Mar20', 'Apr20', 'May20', 'Jun20', 'Jul20',
'Aug20', 'Sep20', 'Oct20', 'Nov20', 'Dec20'
Expected Output:
Id Jan20 Feb20 Mar20 Apr20 May20 Jun20 Jul20 Aug20 Sep20 Oct20 Nov20 Dec20 Amount Activeness
1 20 0 0 12 1 3 1 0 0 2 2 0 100 7
2 0 0 2 1 0 2 0 0 1 0 0 0 500 4
3 1 2 1 2 3 1 1 2 2 3 1 1 300 12
I tried below code:
df['Activeness'] = pd.Series(index=df.index, data=np.count_nonzero(df[['Jan20', 'Feb20',
'Mar20', 'Apr20', 'May20', 'Jun20', 'Jul20',
'Aug20', 'Sep20', 'Oct20', 'Nov20', 'Dec20']], axis=1))
which is working well, but I would like to know is there any method that is faster than this.
You can try:
df['Activeness'] = df.filter(like = '20').ne(0, axis =1).sum(1)
I have a dataframe in pandas, an example of which is provided below:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6
A 1 0 0 1 0 0
B 1 1 0 0 1 0
C 1 0 1 1 0 0
D 0 0 1 0 0 1
E 1 1 1 1 1 1
As you can see 1 and 0 occurs randomly in different columns. It would be helpful, if anyone can suggest me a code in python such that I am able to find the column number where the 1 0 0 pattern occurs for the first time. For example, for member A, the first 1 0 0 pattern occurs at appear_1. so the first occurrence will be 1. Similarly for the member B, the first 1 0 0 pattern occurs at appear_2, so the first occurrence will be at column 2. The resulting table should have a new column named 'first_occurrence'. If there is no such 1 0 0 pattern occurs (like in row E) then the value in first occurrence column will the sum of number of 1 in that row. The resulting table should look something like this:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6 first_occurrence
A 1 0 0 1 0 0 1
B 1 1 0 0 1 0 2
C 1 0 1 1 0 0 4
D 0 0 1 0 0 1 3
E 1 1 1 1 1 1 6
Thank you in advance.
I try not to reinvent the wheel, so I develop on my answer to previous question. From that answer, you need to use additional idxmax, np.where, and get_indexer
cols = ['appear_1', 'appear_2', 'appear_3', 'appear_4', 'appear_5', 'appear_6']
df1 = df[cols]
m = df1[df1.eq(1)].ffill(1).notna()
df2 = df1[m].bfill(1).eq(0)
m2 = df2 & df2.shift(-1, axis=1, fill_value=True)
df['first_occurrence'] = np.where(m2.any(1), df1.columns.get_indexer(m2.idxmax(1)),
df1.shape[1])
Out[540]:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6 first_occurrence
0 A 1 0 0 1 0 0 1
1 B 1 1 0 0 1 0 2
2 C 1 0 1 1 0 0 4
3 D 0 0 1 0 0 1 3
4 E 1 1 1 1 1 1 6
I have a dataset 'df' that looks something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
A 1 0 0 1 0 1
B 1 1 0 0 1 0
C 1 1 1 0 0 1
D 0 0 1 0 0 1
As you can see there are several rows of ones and zeros. Can anyone suggest me a code in python such that I am able to count the number of times '1' occurs continuously before the first occurrence of a 1, 0 and 0 in order. For example, for member A, the first double zero event occurs at seen_2 and seen_3, so the event will be 1. Similarly for the member B, the first double zero event occurs at seen_3 and seen_4 so there are two 1s that occur before this. The resultant table should have a new column 'event' something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
A 1 0 0 1 0 1 1
B 1 1 0 0 1 0 2
C 1 1 1 0 0 1 3
D 0 0 1 0 0 1 1
My approach:
df = df.set_index('MEMBER')
# count 1 on each rows since the last 0
s = (df.stack()
.groupby(['MEMBER', df.eq(0).cumsum(1).stack()])
.cumsum().unstack()
)
# mask of the zeros:
u = s.eq(0)
# look for the first 1 0 0
idx = (~u &
u.shift(-1, axis=1, fill_value=False) &
u.shift(-2, axis=1, fill_value=False) ).idxmax(1)
# look up
df['event'] = s.lookup(idx.index, idx)
Test data:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
0 A 1 0 1 0 0 1
1 B 1 1 0 0 1 0
2 C 1 1 1 0 0 1
3 D 0 0 1 0 0 1
4 E 1 0 1 1 0 0
Output:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
0 A 1 0 1 0 0 1 1
1 B 1 1 0 0 1 0 2
2 C 1 1 1 0 0 1 3
3 D 0 0 1 0 0 1 1
4 E 1 0 1 1 0 0 2
I want the simplest verb that gives a list of all boolean lists of given length.
e.g.
f=. NB. Insert magic here
f 2
0 0
0 1
1 0
1 1
f 3
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
This functionality has been recently added to the stats/base addon.
load 'stats/base/combinatorial' NB. or just load 'stats'
permrep 2 NB. permutations of size 2 from 2 items with replacement
0 0
0 1
1 0
1 1
3 permrep 2 NB. permutations of size 3 from 2 items with replacement
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
permrep NB. display definition of permrep
$:~ :(# #: i.#^~)
Using the Qt IDE you can view the script defining permrep and friends by entering open 'stats/base/combinatorial' in the Term window. Alternatively you can view it on Github.
To define f as specified in your question, the following should suffice:
f=: permrep&2
f=: (# #: i.#^~)&2 NB. alternatively
f 3
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
The #: ("Antibase 2") vocab page has an example close to what I want. I don't really understand that primitive but the following code gives a list of base 2 digits of the numbers 0 to 2^n-1:
f=. #:#i.#(2^])
(Thanks to Dan for getting me to look up #:.)
I'm trying to check the cartesian distance between each set of points in one dataframe to sets of scattered points in another dataframe, to see if the input gets above a threshold 'distance' of my checking points.
I have this working with nested for loops, but is painfully slow (~7 mins for 40k input rows, each checked vs ~180 other rows, + some overhead operations).
Here is what I'm attempting in vectorialized format - 'for every pair of points (a,b) from df1, if the distance to ANY point (d,e) from df2 is > threshold, print "yes" into df1.c, next to input points.
..but I'm getting unexpected behavior from this. With given data, all but one distances are > 1, but only df1.1c is getting 'yes'.
Thanks for any ideas - the problem is probably in the 'df1.loc...' line:
import numpy as np
from pandas import DataFrame
inp1 = [{'a':1, 'b':2, 'c':0}, {'a':1,'b':3,'c':0}, {'a':0,'b':3,'c':0}]
df1 = DataFrame(inp1)
inp2 = [{'d':2, 'e':0}, {'d':0,'e':3}, {'d':0,'e':4}]
df2 = DataFrame(inp2)
threshold = 1
df1.loc[np.sqrt((df1.a - df2.d) ** 2 + (df1.b - df2.e) ** 2) > threshold, 'c'] = "yes"
print(df1)
print(df2)
a b c
0 1 2 yes
1 1 3 0
2 0 3 0
d e
0 2 0
1 0 3
2 0 4
Here is an idea to help you to start...
Source DFs:
In [170]: df1
Out[170]:
c x y
0 0 1 2
1 0 1 3
2 0 0 3
In [171]: df2
Out[171]:
x y
0 2 0
1 0 3
2 0 4
Helper DF with cartesian product:
In [172]: x = df1[['x','y']] \
.reset_index() \
.assign(k=0).merge(df2.assign(k=0).reset_index(),
on='k', suffixes=['1','2']) \
.drop('k',1)
In [173]: x
Out[173]:
index1 x1 y1 index2 x2 y2
0 0 1 2 0 2 0
1 0 1 2 1 0 3
2 0 1 2 2 0 4
3 1 1 3 0 2 0
4 1 1 3 1 0 3
5 1 1 3 2 0 4
6 2 0 3 0 2 0
7 2 0 3 1 0 3
8 2 0 3 2 0 4
now we can calculate the distance:
In [169]: x.eval("D=sqrt((x1 - x2)**2 + (y1 - y2)**2)", inplace=False)
Out[169]:
index1 x1 y1 index2 x2 y2 D
0 0 1 2 0 2 0 2.236068
1 0 1 2 1 0 3 1.414214
2 0 1 2 2 0 4 2.236068
3 1 1 3 0 2 0 3.162278
4 1 1 3 1 0 3 1.000000
5 1 1 3 2 0 4 1.414214
6 2 0 3 0 2 0 3.605551
7 2 0 3 1 0 3 0.000000
8 2 0 3 2 0 4 1.000000
or filter:
In [175]: x.query("sqrt((x1 - x2)**2 + (y1 - y2)**2) > #threshold")
Out[175]:
index1 x1 y1 index2 x2 y2
0 0 1 2 0 2 0
1 0 1 2 1 0 3
2 0 1 2 2 0 4
3 1 1 3 0 2 0
5 1 1 3 2 0 4
6 2 0 3 0 2 0
Try using scipy implementation, it is surprisingly fast
scipy.spatial.distance.pdist
https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.pdist.html
or
scipy.spatial.distance_matrix
https://docs.scipy.org/doc/scipy-0.19.1/reference/generated/scipy.spatial.distance_matrix.html