Python pandas equivalent for replace

Solution 1:

pandas has a replace method too:

In [25]: df = DataFrame({1: [2,3,4], 2: [3,4,5]})

In [26]: df
Out[26]: 
   1  2
0  2  3
1  3  4
2  4  5

In [27]: df[2]
Out[27]: 
0    3
1    4
2    5
Name: 2

In [28]: df[2].replace(4, 17)
Out[28]: 
0     3
1    17
2     5
Name: 2

In [29]: df[2].replace(4, 17, inplace=True)
Out[29]: 
0     3
1    17
2     5
Name: 2

In [30]: df
Out[30]: 
   1   2
0  2   3
1  3  17
2  4   5

or you could use numpy-style advanced indexing:

In [47]: df[1]
Out[47]: 
0    2
1    3
2    4
Name: 1

In [48]: df[1] == 4
Out[48]: 
0    False
1    False
2     True
Name: 1

In [49]: df[1][df[1] == 4]
Out[49]: 
2    4
Name: 1

In [50]: df[1][df[1] == 4] = 19

In [51]: df
Out[51]: 
    1   2
0   2   3
1   3  17
2  19   5

Solution 2:

Pandas doc for replace does not have any examples, so I will give some here. For those coming from an R perspective (like me), replace is basically an all-purpose replacement function that combines the functionality of R functions plyr::mapvalues, plyr::revalue and stringr::str_replace_all. Since DSM covered the case of single values, I will cover the multi-value case.

Example series

In [10]: x = pd.Series([1, 2, 3, 4])

In [11]: x
Out[11]: 
0    1
1    2
2    3
3    4
dtype: int64

We want to replace the positive integers with negative integers (and not by multiplying with -1).

Two lists of values

One way to do this by having one list (or pandas series) of the values we want to replace and a second list with the values we want to replace them with.

In [14]: x.replace([1, 2, 3, 4], [-1, -2, -3, -4])
Out[14]: 
0   -1
1   -2
2   -3
3   -4
dtype: int64

This corresponds to plyr::mapvalues.

Dictionary of value pairs

Sometimes it's more convenient to have a dictionary of value pairs. The index is the one we replace and the value is the one we replace it with.

In [15]: x.replace({1: -1, 2: -2, 3: -3, 4: -4})
Out[15]: 
0   -1
1   -2
2   -3
3   -4
dtype: int64

This corresponds to plyr::revalue.

Strings

It works similarly for strings, except that we also have the option of using regex patterns.

If we simply want to replace strings with other strings, it works exactly the same as before:

In [18]: s = pd.Series(["ape", "monkey", "seagull"])
In [22]: s
Out[22]: 
0        ape
1     monkey
2    seagull
dtype: object

Two lists

In [25]: s.replace(["ape", "monkey"], ["lion", "panda"])
Out[25]: 
0       lion
1      panda
2    seagull
dtype: object

Dictionary

In [26]: s.replace({"ape": "lion", "monkey": "panda"})
Out[26]: 
0       lion
1      panda
2    seagull
dtype: object

Regex

Replace all as with xs.

In [27]: s.replace("a", "x", regex=True)
Out[27]: 
0        xpe
1     monkey
2    sexgull
dtype: object

Replace all ls with xs.

In [28]: s.replace("l", "x", regex=True)
Out[28]: 
0        ape
1     monkey
2    seaguxx
dtype: object

Note that both ls in seagull were replaced.

Replace as with xs and ls with ps

In [29]: s.replace(["a", "l"], ["x", "p"], regex=True)
Out[29]: 
0        xpe
1     monkey
2    sexgupp
dtype: object

In the special case where one wants to replace multiple different values with the same value, one can just simply a single string as the replacement. It must not be inside a list. Replace as and ls with ps

In [29]: s.replace(["a", "l"], "p", regex=True)
Out[29]: 
0        ppe
1     monkey
2    sepgupp
dtype: object

(Credit to DaveL17 in the comments)

Python pandas equivalent for replace

Solution 1:

Solution 2:

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