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Pandas random n samples of consecutive rows / pairs

I have panda dataframe indexed by ID and sorted by value. I want to create a sample size of n=20000 where there are 40000 rows in total and 2 rows are consecutive/paired. I want to perform additional calculations on these 2 consecutive / paired rows
e.g. If I say sample size n=2 I want to randomly pick and find the difference in distance of each of the following picks.
Additional condition: value difference can't exceed 4000.
index value distance
cg13869341 15865 1.635450
cg14008030 18827 4.161332
Then distance of the following etc
cg20826792 29425 0.657369
cg33045430 29407 1.708055
Sample original dataframe
index value distance
cg13869341 15865 1.635450
cg14008030 18827 4.161332
cg12045430 29407 0.708055
cg20826792 29425 0.657369
cg33045430 69407 1.708055
cg40826792 59425 0.857369
cg47454306 88407 0.708055
cg60826792 96425 2.857369
I tried using df_sample = df.sample(n=20000) Then i got bit lost trying to figure out how to get the next row for each value in df_sample
original shape is (480136, 14)

If it doesn't matter to always have (even, odd) pairs (which decreases a bit randomness), you can select n odd rows and get the next even:
N = 20000
# get the indices of N random ODD rows
idx = df.loc[::2].sample(n=N).index
# create a boolean mask to identify the rows
m = df.index.to_series().isin(idx)
# select those OR the next ones
df_sample = df.loc[m|m.shift()]
Example output on the toy DataFrame (N=3):
index value distance
2 cg12045430 29407 0.708055
3 cg20826792 29425 0.657369
4 cg33045430 69407 1.708055
5 cg40826792 59425 0.857369
6 cg47454306 88407 0.708055
7 cg60826792 96425 2.857369
increasing randomness
The drawback of the above approach is that there is a bias to always have (odd, even) pairs. To overcome this we can first remove a random fraction of the DataFrame, small enough to still leave enough choice to pick rows, but large enough to randomly shift the (odd, even) to (even, odd) pairs on many locations. The fraction of rows to remove should be tested depending on the initial size and the sampled size. I used 20-30% here:
N = 20000
frac = 0.2
idx = (df
.drop(df.sample(frac=frac).index)
.loc[::2].sample(n=N)
.index
)
m = df.index.to_series().isin(idx)
df_sample = df.loc[m|m.shift()]
# check:
# len(df_sample)
# 40000

Here's my first attempt (I only just noticed your additional constraint, and I'm not sure if you need the precise number of samples, in which case, you'll have to do some fudging after the line c=c[mask] below).
import random
# Temporarily reset index so we can have something that we can add one to.
df = df.reset_index(level=0)
# Choose the first index of each pair.
# Use random.sample if you don't want repeats,
# or random.choice if you don't mind them.
# The code below does allow overlapping pairs such as (1,2) and (2,3).
first_indices = np.array(random.sample(sorted(df.index[:-1]), 4))
# Filter out those indices where the diff with the next row down is large.
mask = [abs(df.loc[i, "value"] - df.loc[i+1, "value"]) > 4000 for i in c]
c = c[mask]
# Interleave this array with the same numbers, plus 1.
c = np.empty((first_indices.size * 2,), dtype=first_indices.dtype)
c[0::2] = first_indices
c[1::2] = first_indices + 1
# Filter
df_sample = df[df.index.isin(c)]
# Restore original index if required.
df = df.set_index("index")
Hope that helps. Regarding the bit where I use a mask to filter c, this answer might be of help if you need faster alternatives: Filtering (reducing) a NumPy Array

Compute elementwise product for every column in an array efficiently

I have a sparse numpy array with about 60,000 rows and 10,000 columns. Each element in the array is either a 0 or a 1. Each column needs to be multiplied elementwise with four other columns from the matrix and I want to compute all 10,000^5 combinations of columns to multiply. After this step, consecutive 1s in a column are removed such that a 1 can only be followed by a zero. I then multiply this matrix with a vector and save this result (although the code for this is not shown below).
I first tried using numpy with for loops but the run time was too slow so I tried using cupy. It is still very slow. The method I have tried is shifting the arrays in order to compute every possible combination. There are 5 for loops as I want to do elementwise multiplication on 5 arrays. I need all possible combinations of this from the 10,000 columns. The problem is I need to find a faster method and would appreciate any help! Below is the code I have tried so far.
for a in range(1,global_signals.shape[1]):
arr_1 = cp.empty(shape=(global_signals.shape))
arr_1[:,a:] = global_signals[:,:-a]
arr_1[:,:a] = global_signals[:,-a:]
for b in range(1,global_signals.shape[1]):
arr_2 = cp.empty(shape=(global_signals.shape))
arr_2[:,b:] = global_signals[:,:-b]
arr_2[:,:b] = global_signals[:,-b:]
for c in range(1,global_signals.shape[1]):
arr_3 = cp.empty(shape=(global_signals.shape))
arr_3[:,c:] = global_signals[:,:-c]
arr_3[:,:c] = global_signals[:,-c:]
for d in range(1,global_signals.shape[1]):
arr_4 = cp.empty(shape=(global_signals.shape))
arr_4[:,d:] = global_signals[:,:-d]
arr_4[:,:d] = global_signals[:,-d:]
for e in range(1,global_signals.shape[1]):
arr_5 = cp.empty(shape=(global_signals.shape))
arr_5[:,e:] = global_signals[:,:-e]
arr_5[:,:e] = global_signals[:,-e:]
mul_1 = cp.multiply(arr_1,arr_2,arr_3)
mul_2 = cp.multiply(arr_4,arr_5)
mul_result = cp.multiply(mul_1,mul_2)
shifted = cp.append(cp.zeros((1,mul_result.shape[1]),dtype='float32'), mul_result[:-1], axis=0)
final_signals = cp.where((shifted == 0) & (mul_result == 1), 1, 0)
I have recently come to hear of alternative representations for sparse matrices. I would appreciate help on how to best use that, combined with how to best optimise my above code, to make the calculations faster. Please let me know if anything is unclear and needs clarification :)

Pandas apply, rolling, groupby with multiple input & multiple output columns

I’ve been struggling the past week trying to use apply to use functions over an entire pandas dataframe, including rolling windows, groupby, and especially multiple input columns and multiple output columns. I found a large amount of questions on SO about this topic and many old & outdated answers. So I started to create a notebook for every possible combination of x inputs & outputs, rolling, rolling & groupby combined and I focused on performance as well. Since I’m not the only one struggling with these questions I thought I’d provide my solutions here with working examples, hoping it helps any existing/future pandas-users.

Important notes
The combination of apply & rolling in pandas has a very strong output requirement. You have to return one single value. You can not return a pd.Series, not a list, not an array, not secretly an array within an array, but just one value, e.g. one integer. This requirement makes it hard to get a working solution when trying to return multiple outputs for multiple columns. I don’t understand why it has this requirement for 'apply & rolling', because without rolling 'apply' doesn’t have this requirement. Must be due to some internal pandas functions.
The combination of 'apply & rolling' combined with multiple input columns simply does not work! Imagine a dataframe with 2 columns, 6 rows and you want to apply a custom function with a rolling window of 2. Your function should get an input array with 2x2 values - 2 values of each column for 2 rows. But it seems pandas can’t handle rolling and multiple input columns at the same time. I tried to use the axis parameter to get it working but:
Axis = 0, will call your function per column. In the dataframe described above, it will call your function 10 times (not 12 because rolling=2) and since it’s per column, it only provides the 2 rolling values of that column…
Axis = 1, will call your function per row. This is what you probably want, but pandas will not provide a 2x2 input. It actually completely ignores the rolling and only provides one row with values of 2 columns...
When using 'apply' with multiple input columns, you can provide a parameter called raw (boolean). It’s False by default, which means the input will be a pd.Series and thus includes indexes next to the values. If you don’t need the indexes you can set raw to True to get a Numpy array, which often achieves a much better performance.
When combining 'rolling & groupby', it returns a multi-indexes series which can’t easily serve as an input for a new column. The easiest solution is to append a reset_index(drop=True) as answered & commented here (Python - rolling functions for GroupBy object).
You might ask me, when would you ever want to use a rolling, groupby custom function with multiple outputs!? Answer: I recently had to do a Fourier transform with sliding windows (rolling) over a dataset of 5 million records (speed/performance is important) with different batches within the dataset (groupby). And I needed to save both the power & phase of the Fourier transform in different columns (multiple outputs). Most people probably only need some of the basic examples below, but I believe that especially in the Machine Learning/Data-science sectors the more complex examples can be useful.
Please let me know if you have even better, clearer or faster ways to perform any of the solutions below. I'll update my answer and we can all benefit!
Code examples
Let’s create a dataframe first that will be used in all the examples below, including a group-column for the groupby examples.
For the rolling window and multiple input/output columns I just use 2 in all code examples below, but obviously this could be any number > 1.
df = pd.DataFrame(np.random.randint(0,5,size=(6, 2)), columns=list('ab'))
df['group'] = [0, 0, 0, 1, 1, 1]
df = df[['group', 'a', 'b']]
It will look like this:
group a b
0 0 2 2
1 0 4 1
2 0 0 4
3 1 0 2
4 1 3 2
5 1 3 0
Input 1 column, output 1 column
Basic
def func_i1_o1(x):
return x+1
df['c'] = df['b'].apply(func_i1_o1)
Rolling
def func_i1_o1_rolling(x):
return (x[0] + x[1])
df['d'] = df['c'].rolling(2).apply(func_i1_o1_rolling, raw=True)
Roling & Groupby
Add the reset_index solution (see notes above) to the rolling function.
df['e'] = df.groupby('group')['c'].rolling(2).apply(func_i1_o1_rolling, raw=True).reset_index(drop=True)
Input 2 columns, output 1 column
Basic
def func_i2_o1(x):
return np.sum(x)
df['f'] = df[['b', 'c']].apply(func_i2_o1, axis=1, raw=True)
Rolling
As explained in point 2 in the notes above, there isn't a 'normal' solution for 2 inputs. The workaround below uses the 'raw=False' to ensure the input is a pd.Series, which means we also get the indexes next to the values. This enables us to get values from other columns at the correct indexes to be used.
def func_i2_o1_rolling(x):
values_b = x
values_c = df.loc[x.index, 'c'].to_numpy()
return np.sum(values_b) + np.sum(values_c)
df['g'] = df['b'].rolling(2).apply(func_i2_o1_rolling, raw=False)
Rolling & Groupby
Add the reset_index solution (see notes above) to the rolling function.
df['h'] = df.groupby('group')['b'].rolling(2).apply(func_i2_o1_rolling, raw=False).reset_index(drop=True)
Input 1 column, output 2 columns
Basic
You could use a 'normal' solution by returning pd.Series:
def func_i1_o2(x):
return pd.Series((x+1, x+2))
df[['i', 'j']] = df['b'].apply(func_i1_o2)
Or you could use the zip/tuple combination which is about 8 times faster!
def func_i1_o2_fast(x):
return x+1, x+2
df['k'], df['l'] = zip(*df['b'].apply(func_i1_o2_fast))
Rolling
As explained in point 1 in the notes above, we need a workaround if we want to return more than 1 value when using rolling & apply combined. I found 2 working solutions.
1
def func_i1_o2_rolling_solution1(x):
output_1 = np.max(x)
output_2 = np.min(x)
# Last index is where to place the final values: x.index[-1]
df.at[x.index[-1], ['m', 'n']] = output_1, output_2
return 0
df['m'], df['n'] = (np.nan, np.nan)
df['b'].rolling(2).apply(func_i1_o2_rolling_solution1, raw=False)
Pros: Everything is done within 1 function.
Cons: You have to create the columns first and it is slower since it doesn't use the raw input.
2
rolling_w = 2
nan_prefix = (rolling_w - 1) * [np.nan]
output_list_1 = nan_prefix.copy()
output_list_2 = nan_prefix.copy()
def func_i1_o2_rolling_solution2(x):
output_list_1.append(np.max(x))
output_list_2.append(np.min(x))
return 0
df['b'].rolling(rolling_w).apply(func_i1_o2_rolling_solution2, raw=True)
df['o'] = output_list_1
df['p'] = output_list_2
Pros: It uses the raw input which makes it about twice as fast. And since it doesn't use indexes to set the output values the code looks a bit more clear (to me at least).
Cons: You have to create the nan-prefix yourself and it takes a bit more lines of code.
Rolling & Groupby
Normally, I would use the faster 2nd solution above. However, since we're combining groups and rolling this means you'd have to manually set NaN's/zeros (depending on the number of groups) at the right indexes somewhere in the middle of the dataset. To me it seems that when combining rolling, groupby and multiple output columns, the first solution is easier and solves the automatic NaNs/grouping automatically. Once again, I use the reset_index solution at the end.
def func_i1_o2_rolling_groupby(x):
output_1 = np.max(x)
output_2 = np.min(x)
# Last index is where to place the final values: x.index[-1]
df.at[x.index[-1], ['q', 'r']] = output_1, output_2
return 0
df['q'], df['r'] = (np.nan, np.nan)
df.groupby('group')['b'].rolling(2).apply(func_i1_o2_rolling_groupby, raw=False).reset_index(drop=True)
Input 2 columns, output 2 columns
Basic
I suggest using the same 'fast' way as for i1_o2 with the only difference that you get 2 input values to use.
def func_i2_o2(x):
return np.mean(x), np.median(x)
df['s'], df['t'] = zip(*df[['b', 'c']].apply(func_i2_o2, axis=1))
Rolling
As I use a workaround for applying rolling with multiple inputs and I use another workaround for rolling with multiple outputs, you can guess I need to combine them for this one.
1. Get values from other columns using indexes (see func_i2_o1_rolling)
2. Set the final multiple outputs on the correct index (see func_i1_o2_rolling_solution1)
def func_i2_o2_rolling(x):
values_b = x.to_numpy()
values_c = df.loc[x.index, 'c'].to_numpy()
output_1 = np.min([np.sum(values_b), np.sum(values_c)])
output_2 = np.max([np.sum(values_b), np.sum(values_c)])
# Last index is where to place the final values: x.index[-1]
df.at[x.index[-1], ['u', 'v']] = output_1, output_2
return 0
df['u'], df['v'] = (np.nan, np.nan)
df['b'].rolling(2).apply(func_i2_o2_rolling, raw=False)
Rolling & Groupby
Add the reset_index solution (see notes above) to the rolling function.
def func_i2_o2_rolling_groupby(x):
values_b = x.to_numpy()
values_c = df.loc[x.index, 'c'].to_numpy()
output_1 = np.min([np.sum(values_b), np.sum(values_c)])
output_2 = np.max([np.sum(values_b), np.sum(values_c)])
# Last index is where to place the final values: x.index[-1]
df.at[x.index[-1], ['w', 'x']] = output_1, output_2
return 0
df['w'], df['x'] = (np.nan, np.nan)
df.groupby('group')['b'].rolling(2).apply(func_i2_o2_rolling_groupby, raw=False).reset_index(drop=True)

New Dataframe column as a generic function of other rows (pandas)

What is the fastest (and most efficient) way to create a new column in a DataFrame that is a function of other rows in pandas ?
Consider the following example:
import pandas as pd
d = {
'id': [1, 2, 3, 4, 5, 6],
'word': ['cat', 'hat', 'hag', 'hog', 'dog', 'elephant']
}
pandas_df = pd.DataFrame(d)
Which yields:
id word
0 1 cat
1 2 hat
2 3 hag
3 4 hog
4 5 dog
5 6 elephant
Suppose I want to create a new column bar containing a value that is based on the output of using a function foo to compare the word in the current row to the other rows in the dataframe.
def foo(word1, word2):
# do some calculation
return foobar # in this example, the return type is numeric
threshold = some_threshold
for index, _id, word in pandas_df.itertuples():
value = sum(
pandas_df[pandas_df['word'] != word].apply(
lambda x: foo(x['word'], word),
axis=1
) < threshold
)
pandas_df.loc[index, 'bar'] = value
This does produce the correct output, but it uses itertuples() and apply(), which is not performant for large DataFrames.
Is there a way to vectorize (is that the correct term?) this approach? Or is there another better (faster) way to do this?
Notes / Updates:
In the original post, I used edit distance/levenshtein distance as the foo function. I have changed the question in an attempt to be more generic. The idea is that the function to be applied is to compare the current rows value against all other rows and return some aggregate value.
If foo was nltk.metrics.distance.edit_distance and the threshold was set to 2 (as in the original post), this produces the output below:
id word bar
0 1 cat 1.0
1 2 hat 2.0
2 3 hag 2.0
3 4 hog 2.0
4 5 dog 1.0
5 6 elephant 0.0
I have the same question for spark dataframes as well. I thought it made sense to split these into two posts so they are not too broad. However, I have generally found that solutions to similar pandas problems can sometimes be modified to work for spark.
Inspired by this answer to my spark version of this question, I tried to use a cartesian product in pandas. My speed tests indicate that this is slightly faster (though I suspect that may vary with the size of the data). Unfortunately, I still can't get around calling apply().
Example code:
from nltk.metrics.distance import edit_distance as edit_dist
pandas_df2 = pd.DataFrame(d)
i, j = np.where(np.ones((len(pandas_df2), len(pandas_df2))))
cart = pandas_df2.iloc[i].reset_index(drop=True).join(
pandas_df2.iloc[j].reset_index(drop=True), rsuffix='_r'
)
cart['dist'] = cart.apply(lambda x: edit_dist(x['word'], x['word_r']), axis=1)
pandas_df2 = (
cart[cart['dist'] < 2].groupby(['id', 'word']).count()['dist'] - 1
).reset_index()

Let's try to analyze the problem for a second:
If you have N rows, then you have N*N "pairs" to consider in your similarity function. In the general case, there is no escape from evaluating all of them (sounds very rational, but I can't prove it). Hence, you have at least O(n^2) time complexity.
What you can try, however, is to play with the constant factors of that time complexity.
The possible options I found are:
1. Parallelization:
Since you have some large DataFrame, parallelizing the processing is the best obvious choice. That will gain you (almost) linear improvement in time complexity, so if you have 16 workers you will gain (almost) 16x improvement.
For example, we can partition the rows of the df into disjoint parts, and process each part individually, then combine the results.
A very basic parallel code might look like this:
from multiprocessing import cpu_count,Pool
def work(part):
"""
Args:
part (DataFrame) : a part (collection of rows) of the whole DataFrame.
Returns:
DataFrame: the same part, with the desired property calculated and added as a new column
"""
# Note that we are using the original df (pandas_df) as a global variable
# But changes made in this function will not be global (a side effect of using multiprocessing).
for index, _id, word in part.itertuples(): # iterate over the "part" tuples
value = sum(
pandas_df[pandas_df['word'] != word].apply( # Calculate the desired function using the whole original df
lambda x: foo(x['word'], word),
axis=1
) < threshold
)
part.loc[index, 'bar'] = value
return part
# New code starts here ...
cores = cpu_count() #Number of CPU cores on your system
data_split = np.array_split(data, cores) # Split the DataFrame into parts
pool = Pool(cores) # Create a new thread pool
new_parts = pool.map(work , data_split) # apply the function `work` to each part, this will give you a list of the new parts
pool.close() # close the pool
pool.join()
new_df = pd.concat(new_parts) # Concatenate the new parts
Note: I've tried to keep the code as close to OP's code as possible. This is just a basic demonstration code and a lot of better alternatives exist.
2. "Low level" optimizations:
Another solution is to try to optimize the similarity function computation and iterating/mapping. I don't think this will gain you much speedup compared to the previous option or the next one.
3. Function-dependent pruning:
The last thing you can try are similarity-function-dependent improvements. This doesn't work in the general case, but will work very well if you can analyze the similarity function. For example:
Assuming you are using Levenshtein distance (LD), you can observe that the distance between any two strings is >= the difference between their lengths. i.e. LD(s1,s2) >= abs(len(s1)-len(s2)) .
You can use this observation to prune the possible similar pairs to consider for evaluation. So for each string with length l1, compare it only with strings having length l2 having abs(l1-l2) <= limit. (limit is the maximum accepted dis-similarity, 2 in your provided example).
Another observation is that LD(s1,s2) = LD(s2,s1). That cuts the number of pairs by a factor of 2.
This solution may actually get you down to O(n) time complexity (depends highly on the data).
Why? you may ask.
That's because if we had 10^9 rows, but on average we have only 10^3 rows with "close" length to each row, then we need to evaluate the function for about 10^9 * 10^3 /2 pairs, instead of 10^9 * 10^9 pairs. But that's (again) depends on the data. This approach will be useless if (in this example) you have strings all which have length 3.

Thoughts about preprocessing (groupby)
Because you are looking for edit distance less than 2, you can first group by the length of strings. If the difference of length between groups is greater or equal to 2, you do not need to compare them. (This part is quite similar to Qusai Alothman's answer in section 3. H)
Thus, first thing is to group by the length of the string.
df["length"] = df.word.str.len()
df.groupby("length")["id", "word"]
Then, you compute the edit distance between every two consecutive group if the difference in length is less than or equal to 2. This does not directly relate to your question but I hope it would be helpful.
Potential vectorization (after groupby)
After that, you may also try to vectorize the computation by splitting each string into characters. Note that if the cost of splitting is greater than the vectorized benefits it carries, you should not do this. Or when you are creating the data frame, just create one that with characters rather than words.
We will use the answer in Pandas split dataframe column for every character to split a string into a list of characters.
# assuming we had groupped the df.
df_len_3 = pd.DataFrame({"word": ['cat', 'hat', 'hag', 'hog', 'dog']})
# turn it into chars
splitted = df_len_3.word.apply(lambda x: pd.Series(list(x)))
0 1 2
0 c a t
1 h a t
2 h a g
3 h o g
4 d o g
splitted.loc[0] == splitted # compare one word to all words
0 1 2
0 True True True -> comparing to itself is always all true.
1 False True True
2 False True False
3 False False False
4 False False False
splitted.apply(lambda x: (x == splitted).sum(axis=1).ge(len(x)-1), axis=1).sum(axis=1) - 1
0 1
1 2
2 2
3 2
4 1
dtype: int64
Explanation of splitted.apply(lambda x: (x == splitted).sum(axis=1).ge(len(x)-1), axis=1).sum(axis=1) - 1
For each row, lambda x: (x == splitted) compares each row to the whole df just like splitted.loc[0] == splitted above. It will generate a true/false table.
Then, we sum up the table horizontally with a .sum(axis=1) following (x == splitted).
Then, we want to find out which words are similar. Thus, we apply a ge function that checks the number of true is over a threshold. Here, we only allow difference to be 1, so it is set to be len(x)-1.
Finally, we will have to subtract the whole array by 1 because we compare each word with itself in operation. We will want to exclude self-comparison.
Note, this vectorization part only works for within-group similarity checking. You still need to check groups with different length with the edit distance approach, I suppose.

How can I speed up an iterative function on my large pandas dataframe?

I am quite new to pandas and I have a pandas dataframe of about 500,000 rows filled with numbers. I am using python 2.x and am currently defining and calling the method shown below on it. It sets a predicted value to be equal to the corresponding value in series 'B', if two adjacent values in series 'A' are the same. However, it is running extremely slowly, about 5 rows are outputted per second and I want to find a way accomplish the same result more quickly.
def myModel(df):
A_series = df['A']
B_series = df['B']
seriesLength = A_series.size
# Make a new empty column in the dataframe to hold the predicted values
df['predicted_series'] = np.nan
# Make a new empty column to store whether or not
# prediction matches predicted matches B
df['wrong_prediction'] = np.nan
prev_B = B_series[0]
for x in range(1, seriesLength):
prev_A = A_series[x-1]
prev_B = B_series[x-1]
#set the predicted value to equal B if A has two equal values in a row
if A_series[x] == prev_A:
if df['predicted_series'][x] > 0:
df['predicted_series'][x] = df[predicted_series'][x-1]
else:
df['predicted_series'][x] = B_series[x-1]
Is there a way to vectorize this or to just make it run faster? Under the current circumstances, it is projected to take many hours. Should it really be taking this long? It doesn't seem like 500,000 rows should be giving my program that much problem.

Something like this should work as you described:
df['predicted_series'] = np.where(A_series.shift() == A_series, B_series, df['predicted_series'])

df.loc[df.A.diff() == 0, 'predicted_series'] = df.B
This will get rid of the for loop and set predicted_series to the value of B when A is equal to previous A.
edit:
per your comment, change your initialization of predicted_series to be all NAN and then front fill the values:
df['predicted_series'] = np.nan
df.loc[df.A.diff() == 0, 'predicted_series'] = df.B
df.predicted_series = df.predicted_series.fillna(method='ffill')
For fastest speed modifying ayhans answer a bit will perform best:
df['predicted_series'] = np.where(df.A.shift() == df.A, df.B, df['predicted_series'].shift())
That will give you your forward filled values and run faster than my original recommendation

Solution
df.loc[df.A == df.A.shift()] = df.B.shift()