Python: binary vector - python-2.7
I have a set of indices:
indices = (['1', '1.2', '2', '2.2', '3', '4'])
and a dataset, where the first element identifies a person, the second a round, and the third is the index from the indices set:
dataset = [['A', '1', '1'], ['A', '1', '1.2'], ['B', '1', '2'], ['C', '2', '3']]
I would like to form a binary vector, where for each person and for each individual round, the indices are marked either present (with a 1) or not (with a 0).
The desired output would be something like so, where for A, the vector represents the presence of the indices 1 and 1.2, for B, the index 2, and for C, the index 3. Note that for A, there is only one record, but 2 indices are present.
['A', '1', '1, 1, 0, 0, 0, 0']
['B', '1', '0, 0, 1, 0, 0, 0']
['C', '2', '0, 0, 0, 0, 1, 0']
I'm having a bit of trouble getting my head around the looping of the indices over the dataset. My idea was to loop through the indices set the same amount of time as the number of lists in the dataset. But I dont think this is the most efficient way, and any help would be appreciated!
I'd do it something like this:
from itertools import groupby
for k, g in groupby(dataset, lambda x: x[:2]):
vals = [x[2] for x in g]
print(k + [", ".join("1" if x in vals else "0" for x in indices)])
Output
['A', '1', '1, 1, 0, 0, 0, 0']
['B', '1', '0, 0, 1, 0, 0, 0']
['C', '2', '0, 0, 0, 0, 1, 0']
Is this what you were looking for?
Here's a solution without loops
import pandas as pd
indlist=['1', '1.2', '2', '2.2', '3', '4']
dataset = [['A', '1', '1'], ['A', '1', '1.2'], ['B', '1', '2'], ['C', '2', '3']]
df=pd.DataFrame(dataset,columns=['player','round','ind']).set_index('ind').reindex(indlist)
ans=df.reset_index().pivot('player','ind','round').fillna(0)[1:]
Related
Multi-Row Calculation
Suppose I have the following dataframe:
df1 = {'Column_1': ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B'],
'x': ['0', '1', '2', '3', '0', '1', '2', '3']}
df1 = pd.DataFrame (df1, columns = ['Column_1','x'])
df1
I want to create a new column called 'x!'. This is calculated by taking the value in the row 'x' and multiplying it be the row-1 of 'x!'. The value in the first row for 'x!' is 1. I need to calculation to reset when the value in 'Column_1' changes. The desired output would be the follwing:
df2 = {'Column_1': ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B'],
'x': ['0', '1', '2', '3', '0', '1', '2', '3'],
'x!': ['1', '1', '2', '6', '1', '1', '2', '6']}
df2 = pd.DataFrame (df2, columns = ['Column_1','x', 'x!'])
df2
Where 'x' is 3, 'x!' is 6 because 3 x 2 (x! row-1 = 2) is equal to 6.
How would I do this?
Thanks
Try this simple and clear approach:
let df1 = {'Column_1': ['A', 'A', 'A', 'A', 'B', 'B', 'B', 'B'],'x': ['0', '1', '2', '3', '0', '1', '2', '3']}
let newX = ['1'];
for(i=1; i<df1['x'].length; i++){
let check = parseInt(df1.x[i],10)*parseInt(df1.x[i-1],10)
if(check ===0){
newX.push('1')
}else{
newX.push(check.toString())
}
}
console.log(df1.x) //INITIAL ['0', '1', '2', '3', '0', '1', '2', '3']
console.log(newX) //RESULT ['1', '1', '2', '6', '1', '1', '2', '6']
After that you can assign your new x! array:
df1['x!'] = newX
And apply to DataFrame
df2 = pd.DataFrame (df2, columns = Object.keys(df1))
I hope this solves your case. :)
Convert a file into a list with integers
with open('askhsh11.txt', 'r') as f:
raw_list = f.read().split('\n')
for i in range(len(raw_list)):
print raw_list[i].split(',')
for i in range(len(raw_list)):
raw_list[i]=int(i)
print raw_list
the result is :
['1', '2', '3', '4']
['5', '6', '7', '8']
[0, '5,6,7,8']
[0, 1]
but i want the result to be:
['1', '2', '3', '4']
['5', '6', '7', '8']
[1, 2, 3, 4]
[5, 6, 7, 8]
How i convert a list of strings into a integers?
you can use:
result = [int(c) for s in raw_list for c in s.split(',')]
output:
[1, 2, 3, 4, 5, 6, 7, 8]
you can just replace the "'" character to remove them
with open('askhsh11.txt', 'r') as f:
raw_list = f.read().replace("'","").split('\n')
numbers = [int(num) for num in raw_list]
To be honnest . i'm biginner in this langage . but i can tell you that you can use the 2 first output to generate the 2 second output . what i mean that it's easy to convert this : ['1', '2', '3', '4'] to this : [1, 2, 3, 4] . just use the function that convert the string like '1' to the integer 1 . sorry i gorget its name but u find it on python support for evryone who know the function's name .plz make a comment and thank you so much
Subsample rows with conditions in pandas
I am trying to do in pandas something that I can do outside pandas (code below), but it's poorly readable. Goal: subsample rows of a list of lists (or DataFrame) at a maximum interval of 10 rows or less, depending if the values from a "state" column change. In addition, this should be done separately for column values 'a' and 'b' of a dtype column. Code to reproduce the intended output: # input (list of list, but could be converted to DataFrame) # columns: 1:index, 2:state, 3:dtype, 4:value. x = [ [1, 0, 'b', 93.8], [2, 0, 'b', 97.4], [3, 0, 'b', 76.1], [4, 0, 'b', 21.1], [5, 0, 'b', 65.7], [6, 0, 'b', 90.8], [7, 0, 'b', 63.8], [8, 0, 'b', 82.9], [9, 0, 'b', 19.8], [10, 0, 'b', 10.2], [11, 0, 'b', 1.3], [12, 1, 'b', 37.6], [13, 0, 'b', 18.2], [14, 0, 'b', 16.9], [15, 0, 'b', 95.6], [16, 1, 'b', 23.7], [17, 0, 'b', 54.1], [18, 0, 'b', 99.0], [19, 0, 'b', 16.3], [20, 0, 'a', 80.7], [21, 0, 'a', 23.1], [22, 0, 'a', 96.6], [23, 0, 'a', 56.7], [24, 0, 'a', 45.3], [25, 1, 'a', 58.0], [26, 0, 'a', 49.9], [27, 0, 'a', 91.3], [28, 0, 'b', 60.2], [29, 0, 'b', 76.8], [30, 0, 'b', 45.3], [31, 0, 'b', 69.6], [32, 0, 'b', 99.0], [33, 0, 'b', 29.5], [34, 0, 'b', 11.0], [35, 0, 'b', 68.9], [36, 0, 'b', 75.8], [37, 1, 'b', 89.8], [38, 0, 'b', 57.7], [39, 1, 'b', 20.3], [40, 0, 'b', 98.6], [41, 0, 'b', 96.7], [42, 0, 'b', 17.9], [43, 1, 'b', 14.6], [44, 0, 'b', 92.5], [45, 0, 'b', 33.6], [46, 1, 'b', 58.9], [47, 1, 'b', 71.9], [48, 0, 'b', 74.9], [49, 0, 'b', 43.3], [50, 1, 'b', 29.5], [51, 0, 'b', 24.6], [52, 0, 'b', 2.3], [53, 0, 'b', 19.1], [54, 0, 'b', 31.6], [55, 0, 'b', 80.6], [56, 0, 'b', 3.2], [57, 0, 'b', 58.5], [58, 1, 'b', 30.2], [59, 1, 'b', 29.1], [60, 0, 'b', 47.6], [61, 0, 'b', 76.4], [62, 0, 'b', 21.6], [63, 0, 'b', 82.7], [64, 0, 'b', 0.2], [65, 0, 'b', 9.4], [66, 0, 'b', 75.1], [67, 0, 'b', 33.8], [68, 0, 'b', 82.0], [69, 0, 'b', 56.9], [70, 0, 'b', 62.5], [71, 0, 'b', 53.5], [72, 0, 'b', 7.0], [73, 0, 'a', 37.4], [74, 0, 'a', 88.8], [75, 0, 'a', 46.4], [76, 0, 'a', 86.3], [77, 0, 'a', 54.3], [78, 0, 'b', 23.4], [79, 0, 'b', 1.1], [80, 0, 'b', 78.5], [81, 0, 'b', 39.1], [82, 1, 'b', 79.0], [83, 0, 'b', 41.0], [84, 0, 'b', 40.3], [85, 0, 'a', 66.5], [86, 0, 'a', 66.8], [87, 0, 'a', 86.8], [88, 1, 'b', 96.9], [89, 0, 'b', 2.1], [90, 0, 'b', 46.3], [91, 0, 'b', 28.9], [92, 0, 'b', 43.2], [93, 0, 'b', 58.9], [94, 0, 'b', 60.6], [95, 0, 'b', 15.4], [96, 0, 'b', 69.4], [97, 1, 'b', 18.4], [98, 0, 'b', 41.3], [99, 0, 'b', 40.5] ] ] Code to resample x for state 'a' and 'b': def resample(x, log_interval, dtype): if not x: return red = [] prev_state, next_val, last_val = 0, 0, 0 for row in x: if row[2] == dtype: if row[0] >= next_val or row[1] != prev_state and row[0] > last_val: red.append(row) prev_state = row[1] next_val = row[0] + log_interval last_val = row[0] return red red_a = resample(x, 10, 'a') red_b = resample(x, 10, 'b') And expected outcome for red_a and red_b: red_a = [ [20, 0, a, 80.7], [25, 1, a, 58.0], [26, 0, a, 49.9], [73, 0, a, 37.4], [85, 0, a, 66.5] ] red_b = [ [1, 0, b, 93.8], [11, 0, b, 1.3], [12, 1, b, 37.6], [13, 0, b, 18.2], [16, 1, b, 23.7], [17, 0, b, 54.1], [28, 0, b, 60.2], [37, 1, b, 89.8], [38, 0, b, 57.7], [39, 1, b, 20.3], [40, 0, b, 98.6], [43, 1, b, 14.6], [44, 0, b, 92.5], [46, 1, b, 58.9], [48, 0, b, 74.9], [50, 1, b, 29.5], [51, 0, b, 24.6], [58, 1, b, 30.2], [60, 0, b, 47.6], [70, 0, b, 62.5], [80, 0, b, 78.5], [82, 1, b, 79.0], [83, 0, b, 41.0], [88, 1, b, 96.9], [89, 0, b, 2.1], [97, 1, b, 18.4], [98, 0, b, 41.3] ] How can I do this in pandas? A starting point is: columns = ['ind', 'state', 'dtype', 'value'] df = pd.DataFrame(x, columns=columns) But if I try a for loop it is extremely slow (eg for row in df: ...). Any idea how to proceed from here?
So starting with df = pd.DataFrame(x, columns=['ind', 'state', 'dtype', 'value']), at first you can create two DFs (df_a and df_b) selecting the states such as:
df_a = df[df['dtype'] =='a'].copy()
df_b = df[df['dtype'] =='b'].copy()
Then you create a function select_row that you will apply to these DFs:
def select_row( row, log_interval):
# using global varaibles might be a bit dangerous but I didn't find another way
global prev_state, next_val, last_val
# Here your conditions
if (row['ind'] >= next_val) or (row['state'] != prev_state and row['ind'] > last_val):
# change the values of the global variables
prev_state = row['state']
next_val = row['ind'] + log_interval
last_val = row['ind']
return True # return True if your condition is met
else: # return False otherwise
return False
Now you can create a column in df_a and df_b with a Boolean value such as:
log_interval = 10
prev_state, next_val, last_val = 0, 0, 0
df_a['bool'] = df_a.apply(select_row, args = ([log_interval ]), axis = 1)
#same for df_b but don't forget to reset your global values
prev_state, next_val, last_val = 0, 0, 0
df_b['bool'] = df_b.apply(select_row, args = ([log_interval ]), axis = 1)
Finally, you can create your two output by selecting the row of df_a (and df_b) having True in column 'bool' and drop this column:
red_a = df_a[df_a['bool'] == True].drop('bool',axis=1)
red_b = df_b[df_b['bool'] == True].drop('bool',axis=1)
Integrate multiple dictionaries in django
Really having trouble about list these days, I'm a developer from php looking a place in python. My question is in relation to my previous question
I now have a dictionary group by id_position and flag that contains order [Top, Right, Bottom, Left, Center]:
a = {'41': [0, 0, 0, 0, 1], '42': [0, 0, 1, 0, 1], '43': [0, 0, 0, 0, 1], '44': [0, 0, 0, 0, 1]}
and other dictionary that contains my id_position and status:
b = {'44': 'statusC', '42': 'statusB', '41': 'statusA', '43': 'statusC'}
I want to include dict A in my code to save dict B below.
for pos, stat in B.items():
MyModel.objects.create(position=pos, status=stat, Top = "" , Right="" Bottom = "", Left= "")
How can I make this wok?
Can you recommend study list where I can start to work from php to django.
UPDATE:
I followed this and added my code below:
c = {}
for key in set().union(a, b):
if key in a: c.setdefault(key, []).extend(a[key])
if key in b: c.setdefault(key, []).extend(b[key])
print(c)
and it returned:
{
'42': [0, 0, 1, 0, 1, 's', 't', 'a', 't', 'u', 's', 'B'],
'41': [0, 0, 0, 0, 1, 's', 't', 'a', 't', 'u', 's', 'A'],
'44': [0, 0, 0, 0, 1, 's', 't', 'a', 't', 'u', 's', 'C'],
'43': [0, 0, 0, 0, 1, 's', 't', 'a', 't', 'u', 's', 'C']
}
my problem now is that my string is being separated
Not sure what you're looking for... something like this?: Top=A[int(pos)][0] Right=A[int(pos)][1] Bottom=A[int(pos)][2] Left=A[int(pos)][3] hope it helps.
Find overlapping List in 2 Lists of Lists
I would like to find overlapping lists in two lists of lists. ListLeft = [['A', 'B', 'C'], ['1', '2', '3', '4'], ['x', 'y'], ['one', 'two', 'three']] ListRight = [['h', 'i', 'j'], ['A', 'B', 'C'], ['1', '2', '3', '4'], ['5', '6', '7'], ['x', 'y']] Someone might have a solution to find/print content of overlapping lists and lists which are not in both lists Maybe this is possible without importing modules.
This can be simply achieved by using loop: overlap = [] for ll in ListLeft: for lr in ListRight: if ll == lr: overlap.append(ll) break print overlap >>> [['A', 'B', 'C'], ['1', '2', '3', '4'], ['x', 'y']]