Find and replace multiple values in python

问题:

I want to find and replace multiple values in an 1D array / list with new ones.

In example for a list

a=[2, 3, 2, 5, 4, 4, 1, 2]

I would like to replace

val_old=[1, 2, 3, 4, 5] 

with

val_new=[2, 3, 4, 5, 1]

Therefore the new array is:

a_new=[3, 4, 3, 1, 5, 5, 2, 3]

What is the fastest way to do this (for very large lists, i.e. with 50000 values to find and replace)?

Comment of the anwsers

Thank you to all for a quick response! I checked the proposed solutions with the following:

N = 10**4
N_val = 0.5*N
a = np.random.randint(0, N_val, size=N)
val_old = np.arange(N_val, dtype=np.int)
val_new = np.arange(N_val, dtype=np.int)
np.random.shuffle(val_new)

a1 = list(a)
val_old1 = list(val_old)
val_new1 = list(val_new)

def Ashwini_Chaudhary(a, val_old, val_new):
    arr = np.empty(a.max()+1, dtype=val_new.dtype)
    arr[val_old] = val_new
    return arr[a]

def EdChum(a, val_old, val_new):
    df = pd.Series(a, dtype=val_new.dtype)
    d = dict(zip(val_old, val_new))
    return df.map(d).values   

def xxyzzy(a, val_old, val_new):
    return [val_new[val_old.index(x)] for x in a]

def Shashank_and_Hackaholic(a, val_old, val_new):
    d = dict(zip(val_old, val_new))
    return [d.get(e, e) for e in a]

def itzmeontv(a, val_old, val_new):
    return [val_new[val_old.index(i)] if i in val_old else i for i in a]

def swenzel(a, val_old, val_new):
    return val_new[np.searchsorted(val_old,a)]

def Divakar(a, val_old, val_new):
    C,R = np.where(a[:,np.newaxis] == val_old[np.newaxis,:])
    a[C] = val_new[R]
    return a

The results:

%timeit -n100 Ashwini_Chaudhary(a, val_old, val_new)
100 loops, best of 3: 77.6 µs per loop

%timeit -n100 swenzel(a, val_old, val_new)
100 loops, best of 3: 703 µs per loop

%timeit -n100 Shashank_and_Hackaholic(a1, val_old1, val_new1)
100 loops, best of 3: 1.7 ms per loop

%timeit -n100 EdChum(a, val_old, val_new)
100 loops, best of 3: 17.6 ms per loop

%timeit -n10 Divakar(a, val_old, val_new)
10 loops, best of 3: 209 ms per loop

%timeit -n10 xxyzzy(a1, val_old1, val_new1)
10 loops, best of 3: 429 ms per loop

%timeit -n10 itzmeontv(a1, val_old1, val_new1)
10 loops, best of 3: 847 ms per loop

The relative difference in performance increases with biger N , i.e. if N=10**7, then the result by Ashwini_Chaudhary takes 207 ms and the result by swenzel 6.89 s.

回答1:

>>> arr = np.empty(a.max() + 1, dtype=val_new.dtype)
>>> arr[val_old] = val_new
>>> arr[a]
array([3, 4, 3, 1, 5, 5, 2, 3])

回答2:

In vanilla Python, without the speed of numpy or pandas, this is one way:

a = [2, 3, 2, 5, 4, 4, 1, 2]
val_old = [1, 2, 3, 4, 5]
val_new = [2, 3, 4, 5, 1]
expected_a_new = [3, 4, 3, 1, 5, 5, 2, 3]
d = dict(zip(val_old, val_new))
a_new = [d.get(e, e) for e in a]
print a_new # [3, 4, 3, 1, 5, 5, 2, 3]
print a_new == expected_a_new # True

The average time complexity for this algorithm is O(M + N) where M is the length of your "translation list" and N is the length of list a.

回答3:

Assuming that your val_old array is sorted (which is the case here, but if later on it's not, then don't forget to sort val_new along with it!), you can use numpy.searchsorted and then access val_new with the results.
This does not work if a number has no mapping, you will have to provide 1to1 mappings in that case.

In [1]: import numpy as np

In [2]: a = np.array([2, 3, 2, 5, 4, 4, 1, 2])

In [3]: old_val = np.array([1, 2, 3, 4, 5])

In [4]: new_val = np.array([2, 3, 4, 5, 1])

In [5]: a_new = np.array([3, 4, 3, 1, 5, 5, 2, 3])

In [6]: i = np.searchsorted(old_val,a)

In [7]: a_replaced = new_val[i]

In [8]: all(a_replaced == a_new)
Out[8]: True

50k numbers? No problem!

In [23]: def timed():
    t0 = time.time()
    i = np.searchsorted(old_val, a)
    a_replaced = new_val[i]
    t1 = time.time()
    print('%s Seconds'%(t1-t0))
   ....: 

In [24]: a = np.random.choice(old_val, 50000)

In [25]: timed()
0.00288081169128 Seconds

500k? You won't notice the difference!

In [26]: a = np.random.choice(old_val, 500000)

In [27]: timed()
0.019248008728 Seconds

回答4:

Try this for your expected output, works even if elements not in value_old.

>>>[val_new[val_old.index(i)] if i in val_old else i for i in a]
[3, 4, 3, 1, 5, 5, 2, 3]

回答5:

The numpy_indexed package (disclaimer: I am its author) provides an elegant and efficient vectorized solution to this type of problem:

import numpy_indexed as npi
remapped_a = npi.remap(a, val_old, val_new)

The method implemented is based on searchsorted like that of swenzel and should have similar good performance, but more general. For instance, the items of the array do not need to be ints, but can be any type, even nd-subarrays themselves.

If all values in 'a' are expected to be present in 'val_old', you can set the optional 'missing' kwarg to 'raise' (default is 'ignore'). Performance will be slightly better, and you will get a KeyError if that assumption is not satisfied.

回答6:

To replace values in a list using two other lists as key:value pairs there are several approaches. All of them use "list compression".

Using list.index():

a=[2, 3, 2, 5, 4, 4, 1, 2]
val_old=[1, 2, 3, 4, 5] 
val_new=[2, 3, 4, 5, 1]
a_new=[val_new[val_old.index(x)] for x in a]

Using your special case:

a=[2, 3, 2, 5, 4, 4, 1, 2]
a_new=[x % 5 + 1 for x in a]

回答7:

I tried like this:

>>> val_old=[1, 2, 3, 4, 5]
>>> val_new=[2, 3, 4, 5, 1]
>>> a=[2, 3, 2, 5, 4, 4, 1, 2]
>>> my_dict = dict(zip(val_old, val_new))
>>> [my_dict.get(x,x) for x in a]
[3, 4, 3, 1, 5, 5, 2, 3]

回答8:

In pandas I would create a dict from the 2 lists and then call map which will perform a lookup and replace the values:

In [6]:

df = pd.Series([2, 3, 2, 5, 4, 4, 1, 2])
df
Out[6]:
0    2
1    3
2    2
3    5
4    4
5    4
6    1
7    2
dtype: int64
In [7]:

val_old=[1, 2, 3, 4, 5] 
val_new=[2, 3, 4, 5, 1]
d = dict(zip(val_old,val_new ))
d
Out[7]:
{1: 2, 2: 3, 3: 4, 4: 5, 5: 1}
In [9]:

df.map(d)

Out[9]:
0    3
1    4
2    3
3    1
4    5
5    5
6    2
7    3
dtype: int64

For a 80000 element series this takes 3.4ms:

In [14]:

%timeit df.map(d)

100 loops, best of 3: 3.4 ms per loop

This is a vectorised approach and will scale much better than any iteration based method

回答9:

For numpy arrays, this could be one approach -

%// Find row and column IDs for matches between "a" and "val_old"
C,R = np.where(a[:,np.newaxis] == val_old[np.newaxis,:])

%// Index into "a" with the column indices and 
%// set those to "val_new" elements indexed by "R"
a[C] = val_new[R]

---

Sample run and timing

For inputs:

a = np.random.randint(10000,size=(100000))
val_old = np.random.randint(10000,size=(1000))
val_new = np.random.randint(10000,size=(1000))

Runtimes at each code line were -

%timeit C,R = np.where(a[:,np.newaxis] == val_old[np.newaxis,:])
1 loops, best of 3: 292 ms per loop

%timeit a[C] = val_new[R]
10000 loops, best of 3: 43 µs per loop

回答10:

list(map(lambda x:val_new[val_old.index(x)], a))