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Given a list of string,
['a', 'a', 'c', 'a', 'a', 'a', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'b', 'b', 'b', 'd', 'b', 'b', 'b']
I would like to convert to an integer-category form
[0, 0, 2, 0, 0, 0, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 1, 1, 1, 3, 1, 1, 1]
This can achieve using numpy unique as below
ipt=['a', 'a', 'c', 'a', 'a', 'a', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'b', 'b', 'b', 'd', 'b', 'b', 'b']
_, opt = np.unique(np.array(ipt), return_inverse=True)
But, I curious if there is another alternative without the need to import numpy.
If you are solely interested in finding integer representation of factors, then you can use a dict comprehension along with enumerate to store the mapping, after using set to find unique values:
lst = ['a', 'a', 'c', 'a', 'a', 'a', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'b', 'b', 'b', 'd', 'b', 'b', 'b']
d = {x: i for i, x in enumerate(set(lst))}
lst_new = [d[x] for x in lst]
print(lst_new)
# [3, 3, 0, 3, 3, 3, 2, 0, 2, 2, 2, 2, 0, 2, 2, 2, 2, 0, 2, 2, 2, 2, 0, 1, 1, 1, 2, 1, 1, 1]
This approach can be used for general factors, i.e., the factors do not have to be 'a', 'b' and so on, but can be 'dog', 'bus', etc. One drawback is that it does not care about the order of factors. If you want the representation to preserve order, you can use sorted:
d = {x: i for i, x in enumerate(sorted(set(lst)))}
lst_new = [d[x] for x in lst]
print(lst_new)
# [0, 0, 2, 0, 0, 0, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 1, 1, 1, 3, 1, 1, 1]
You could take a note out of the functional programming book:
ipt=['a', 'a', 'c', 'a', 'a', 'a', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'd', 'd', 'd', 'd', 'c', 'b', 'b', 'b', 'd', 'b', 'b', 'b']
opt = list(map(lambda x: ord(x)-97, ipt))
This code iterates through the input array and passes each element through the lambda function, which takes the ascii value of the character, and subtracts 97 (to convert the characters to 0-25).
If each string isn't a single character, then the lambda function may need to be adapted.
You could write a custom function to do the same thing as you are using numpy.unique() for.
def unique(my_list):
''' Takes a list and returns two lists, a list of each unique entry and the index of
each unique entry in the original list
'''
unique_list = []
int_cat = []
for item in my_list:
if item not in unique_list:
unique_list.append(item)
int_cat.append(unique_list.index(item))
return unique_list, int_cat
Or if you wanted your indexing to be ordered.
def unique_ordered(my_list):
''' Takes a list and returns two lists, an ordered list of each unique entry and the
index of each unique entry in the original list
'''
# Unique list
unique_list = []
for item in my_list:
if item not in unique_list:
unique_list.append(item)
# Sorting unique list alphabetically
unique_list.sort()
# Integer category list
int_cat = []
for item in my_list:
int_cat.append(unique_list.index(item))
return unique_list, int_cat
Comparing the computation time for these two vs numpy.unique() for 100,000 iterations of your example list, we get:
numpy = 2.236004s
unique = 0.460719s
unique_ordered = 0.505591s
Showing that either option would be faster than numpty for simple lists. More complicated strings decrease the speed of unique() and unique_ordered much more than numpy.unique(). Doing 10,000 iterations of a random, 100 element list of 20 character strings, we get times of:
numpy = 0.45465s
unique = 1.56963s
unique_ordered = 1.59445s
So if efficiency was important and your list had more complex/a larger variety of strings, it would likely be better to use numpy.unique()
I have some list A = ['a', 'b', 'c', 'd', 'e', 'f']
I need take indices of elements in this order 0 1 1 2 2 3 3 4 4 5
But my code made this order 0 1 2 3 4 5
A = ['a', 'b', 'c', 'd', 'e', 'f']
for i in A:
print(A.index(i), end=' ')
If you have the desired indices why not try this:
X = [0, 1, 1, 2, 2, 3, 3, 4, 4, 5]
A = ['a', 'b', 'c', 'd', 'e', 'f']
for i in X:
print(A[i], end=' ')
Using list comprehension to extract the values corresponding to the indices.
X = [0, 1, 1, 2, 2, 3, 3, 4, 4, 5]
A = ['a', 'b', 'c', 'd', 'e', 'f']
new_list = [A[x] for x in X]
Update
How to make a flat list out of a list of lists Used to flatten nested list
list_of_list = [[x,x] for x in range(1,len(A))]
new_list = [0]+[item for sublist in list_of_list for item in sublist]
I have a following problem. Suppose I have this dataframe:
import pandas as pd
d = {'Name': ['c', 'c', 'c', 'a', 'a', 'b', 'b', 'd', 'd'], 'Project': ['aa','ab','bc', 'aa', 'ab','aa', 'ab','ca', 'cb'],
'col2': [3, 4, 0, 6, 45, 6, -3, 8, -3]}
df = pd.DataFrame(data=d)
I need to add a new column that add a number to each project per name. Desired output is:
import pandas as pd
dnew = {'Name': ['c', 'c', 'c', 'a', 'a', 'b', 'b', 'd', 'd'], 'Project': ['aa','ab','bc', 'aa', 'ab','aa', 'ab','ca', 'cb'],
'col2': [3, 4, 0, 6, 45, 6, -3, 8, -3], 'New_column': ['1', '1','1','2', '2','2','2','3','3']}
NEWdf = pd.DataFrame(data=dnew)
In other words: 'aa','ab','bc' in Project occurs in the first rows, so I add 1 to the new column. 'aa', 'ab' is the second Project from the beginning. It occurs for Name 'a' and 'b', so I add 2 to the both new column. 'ca', 'cb' is the third project and it occurs only for name 'd', so I add 3 only to the name 'd'.
I tried to combine groupby with a for loop, but it did not worked to me. Thanks a lot for a help!
Looks like networkx since Name and Project are related , you can use:
import networkx as nx
G=nx.from_pandas_edgelist(df, 'Name', 'Project')
l = list(nx.connected_components(G))
s = pd.Series(map(list,l)).explode()
df['new'] = df['Project'].map({v:k for k,v in s.items()}).add(1)
print(df)
Name Project col2 new
0 a aa 3 1
1 a ab 4 1
2 b bb 6 2
3 b bc 6 2
4 c aa 6 1
5 c ab 6 1
Suppose I have two datasets
DS1
ArrayCol
[1,2,3,4]
[1,2,3]
DS2
Key Name
1 A
2 B
3 C
4 D
how to look up the values in the array to map the "Name" so that I can have another dataset like the following?
DS3
COlNew
[A,B,C,D]
[A,B,C]
Thanks, it's in databricks, so method is ok . python,sql,scala…...
you can try this
ds1 = [[1, 2, 3, 4], [1, 2, 3]]
ds2 = {1: 'A', 2: 'B', 3: 'C', 4: 'D'}
new_data = [[ds2[cell] for cell in col] for col in ds1]
print(new_data)
output:
[['A', 'B', 'C', 'D'], ['A', 'B', 'C']]
hope that will be help. :)
Lets consider your dataset are in files and you can do something like this,
making use of dict
f=open("ds1.txt").readlines()
g=open("ds2.txt").readlines()
u=dict(item.rstrip().split("\t") for item in g)
for i in f:
i = i.rstrip().strip('][').split(',')
print [u[col] for col in i]
Output
['A', 'B', 'C', 'D']
['A', 'B', 'C']
I have pandas dataframe that looks like this:
df = pd.DataFrame({'name': [0, 1, 2, 3], 'cards': [['A', 'B', 'C', 'D'],
['B', 'C', 'D', 'E'],
['E', 'F', 'G', 'H'],
['A', 'A', 'E', 'F']]})
name cards
0 ['A', 'B', 'C', 'D']
1 ['B', 'C', 'D', 'E']
2 ['E', 'F', 'G', 'H']
3 ['A', 'A', 'E', 'F']
And I'd like to create a matrix that looks like this:
name 0 1 2 3
name
0 4 3 0 1
1 3 4 1 1
2 0 1 4 2
3 1 1 2 4
Where the values are the number of items in common.
Any ideas?
Using .apply method and lambda we can directly get a dataframe
def func(df, j):
return pd.Series([len(set(i)&set(j)) for i in df.cards])
newdf = df.cards.apply(lambda x: func(df, x))
newdf
0 1 2 3
0 4 3 0 1
1 3 4 1 1
2 0 1 4 2
3 1 1 2 3
By list comprehension and iterate through all pairs we can make the result:
import pandas as pd
df = pd.DataFrame({'name': [0, 1, 2, 3], 'cards': [['A', 'B', 'C', 'D'],
['B', 'C', 'D', 'E'],
['E', 'F', 'G', 'H'],
['A', 'A', 'E', 'F']]})
result=[[len(list(set(x) & set(y))) for x in df['cards']] for y in df['cards']]
print(result)
output :
[[4, 3, 0, 1], [3, 4, 1, 1], [0, 1, 4, 2], [1, 1, 2, 3]]
'&' is used to calculate intersection of two sets
This is exactly what you want:
import pandas as pd
df = pd.DataFrame({'name': [0, 1, 2, 3], 'cards': [['A', 'B', 'C', 'D'],
['B', 'C', 'D', 'E'],
['E', 'F', 'G', 'H'],
['A', 'A', 'E', 'F']]})
result=[[len(x)-max(len(set(y) - set(x)),len(set(x) - set(y))) for x in df['cards']] for y in df['cards']]
print(result)
output:
[[4, 3, 0, 1], [3, 4, 1, 1], [0, 1, 4, 2], [1, 1, 2, 4]]
import pandas as pd
import numpy as np
df = pd.DataFrame([['A', 'B', 'C', 'D'],
['B', 'C', 'D', 'E'],
['E', 'F', 'G', 'H'],
['A', 'A', 'E', 'F']])
nrows = df.shape[0]
# Initialization
matrix = np.zeros((nrows,nrows),dtype= np.int64)
for i in range(0,nrows):
for j in range(0,nrows):
matrix[i,j] = sum(df.iloc[:,i] == df.iloc[:,j])
output
print(matrix)
[[4 1 0 0]
[1 4 0 0]
[0 0 4 0]
[0 0 0 4]]