When regex=True, value is not None and to_replace is a string, the replacement will be applied in all columns of the DataFrame. >>> df = pd.DataFrame( ... { ... "A": [0, 1, 2, 3, 4], ... "B": ["a", "b", "c", "d", "e"], ... "C": ["f", "g", "h", "i", "j"], ...

February 10, 2025 - Let’s take a look at how replace() works in different situations: ... If you want to replace one specific value with another, you can just pass those values directly into the function. For example, if you have a column with the value 3 that needs to be changed to 10, here's how you can do it: import pandas as pd df = pd.DataFrame({'A': [1, 2, 3, 4, 5]}) df = df.replace(3, 10) print(df)

March 2, 2023 - However, it’s not my preferred approach as the behavior can often be difficult to read. Let’s take a look at how the method can replace values: # Using a Dictionary (Dict is passed into to_replace=) df['Age'] = df['Age'].replace({23:99, 45:999}) # Using a Dictionary for Column Replacements (key:value = column:value) df = df.replace({'Name': 'Jane', 'Age': 45}, 99) We can see that the dictionary can be used in two different ways: To map values to replace so that the dictionary represents {original value : new value} To map replacements from columns so that it follows the structure shown here: to_replace={column1: value1, column2: value2}, value=new value ·

May 1, 2024 - I tried this on a dataframe df that has one column containing the strings “A” and “B”. I want to change B to 1 and A to 0. Here is how I tried: # Define the mapping dictionary mapping = {'A': 0, 'B': 1} # Replace values in the entire DataFrame using the mapping dictionary df.replace(mapping, inplace=True) print(df) But when I look at the entries in that (after exporting to a csv so I can read it), it almost does the job.

March 11, 2017 - IndexingRelated to indexing on series/frames, not to indexes themselvesRelated to indexing on series/frames, not to indexes themselvesUsage Question ... # Your code here data = pd.DataFrame( {'ab' : ['A','B','A','A','B'], 'num' : ['01','02','01','01','01']}) a_replacements = { 'num' : { '01' : 'funny', '02' : 'serious' }} b_replacements = { 'num' : { '01' : 'beginning', '02' : 'end' }} data[data.ab == 'A'].replace(inplace=True, to_replace=a_replacements)

December 27, 2024 - Apply regex within replace() for pattern-based replacement. ... df = pd.DataFrame({ 'A': ['foo', 'bar', 'baz'], 'B': ['foobar', 'barfoo', 'foobarbaz'] }) df.replace(r'^foo', 'new', regex=True) Explain Code

May 13, 2023 - To replace a character in df in Python we can: df.replace("?", np.nan, inplace = True) but it seems Julia not support replace to the whole DataFrame replace(df, "?" => missing) returns: MethodError: no method matching similar(::DataFrames.DataFrame, ::Type{Any}) After some testing, I found the alternative using for loop: for col in names(df) replace!(df[!, col], "?" => missing) end but actually I prefer one-liner code than for loop.

November 17, 2021 - Also please note both of the following work fine: import pandas as pd df = pd.DataFrame({"a": ["x", "y", "z", "w"], "b": [True, False, False, True]}).astype({"b":"bool"}) # notice 'bool' instead of 'boolean' df.replace("x", "X") and: df = pd.DataFrame({"a": ["x", "y", "z", "w"], "b": [True, False, False, True]}) df.replace("x", "X") the replace should complete successfully ·

August 4, 2023 - I have checked that this issue has not already been reported. I have confirmed this bug exists on the latest version of pandas. I have confirmed this bug exists on the main branch of pandas. import pandas as pd # FAILS df1 = pd.DataFrame({ "a": pd.Series([1], dtype="int64"), "b": pd.Series([None], dtype="object"), }) # PASSES df2 = pd.DataFrame({ "a": [1], "b": [None], }, dtype="object") # PASSES df3 = pd.DataFrame({ "a": pd.Series([1], dtype="object"), "b": pd.Series([None], dtype="object"), }) # PASSES df4 = pd.DataFrame({ "a": pd.Series([1], dtype="int64"), "b": pd.Series([None], dtype="obj

April 1, 2022 - I have checked that this issue has not already been reported. I have confirmed this bug exists on the latest version of pandas. I have confirmed this bug exists on the main branch of pandas. import pandas as pd df = pd.DataFrame(dict(a=[1,2,3], b=[1,2,3])) df.replace({1:5}, value=None) # does not replace values at all df.replace({1:5}) # correctly replaces 1s with 5s

April 12, 2021 - Hi, I am using Pandas' replace function and it works on one DataFrame but not on another and I don't understand why. I have tried many different solutions but nothing seems to work - Please take a look at the example code below: import pandas as p...

January 14, 2022 - import pandas as pd import numpy as np list_of_random_numbers = np.random.randint(0,100,size=10) list_of_random_numbers_as_string = [str(x) for x in np.random.randint(0,100,size=10)] list_of_random_numbers_as_string[0] = list_of_random_numbers_as_string[0] + "'" df = pd.concat( [pd.DataFrame({'value':list_of_random_numbers}), pd.DataFrame({'value':list_of_random_numbers_as_string})], ignore_index=True) df['value'] = df['value'].str.replace("'", "") df.notnull().sum() When replacing a string in a column with mixed data, this replaces all non-string data with NaN without a warning (silent failure).

April 11, 2019 - import pandas as pd import numpy as np ar = np.random.normal(size=[100,10]) df = pd.DataFrame(ar).astype(str) df.replace('0',None) Every time I run something similar to this (replacing, in a string DataFrame/DataFrame column, a specific string with None), the columns get filled with the element directly above it in the column; this creates a lot of confusion since None is, pretty much, THE "pythonic" way of signalling the absence of a variable/whatever. It works fine if replaced with np.nan, but if anything, I would expect them to work very approximately the same.

With regex=True it will replace commas contained within cells with other text. This is the standard way to solve this problem at the DataFrame level. ... Save this answer. ... Show activity on this post.