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I am still a beginner in python. I am trying to find a common value with if statement,

import pandas as pd

df = pd.read_csv("data.csv")

for n in range(2, len(df)):
    if df.loc[n].isin([2]).any():
        if df.loc[n-1].isin([1]).any():
            print("Yes")

n += 1
A B
1 3
2 3
1 8
1 2
3 5
2 7

I want it to be printed "Yes" if every time 2 is in n and 1 is in n-1 until the end of data. not just if it matches once. If ever 2 is in n and n-1 doesn't have any value 1, then it won't be printed. For example the above table won't be printed as not all 2 in n doesn't have 1 in n-1

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  • 1
    What's with the n += 1 at the end? If that's there by accident, please edit to remove it. Commented Nov 4 at 17:01
  • 1
    What's the desired output exactly, and for which rows? Please edit to clarify. You can also remove pd.read_csv("data.csv") since that's not relevant to the question; instead it would be better to provide the data in the code. See How to make good reproducible pandas examples. (See also minimal reproducible example.) Commented Nov 4 at 17:10
  • Why are you using a loop in the first place? Are you aware that iteration in Pandas is an anti-pattern? Maybe you don't know how to do the equivalent of n-1? In that case, I can show you. Commented Nov 4 at 17:10
  • 1
    Maybe the simplest way to clarify the output would be to add print("No") for non-matching rows. Speaking of that, there's no need to nest ifs here when you could use and instead. That would save you one level of nesting. Commented Nov 4 at 17:17
  • 1
    After reading the answer, it seems like I totally misunderstood the output you want because your description is unclear. You meant to say "if every time 2 is in n then 1 is in n-1". Commented Nov 4 at 17:21

3 Answers 3

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1

i would give a try to this code:

res = (df.loc[df["A"]==2].index-1).isin(df.loc[df["A"]==1].index).all()

then print your dataframe bases on boolean value in res.

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1

You can solve this efficiently with a vector operation. You need to find:

  • all rows in which there is at least one "2" with eq + any
  • access the previous rows with shift
  • test again for those if any value is "1"
  • check if all such rows are True
out = df.shift()[df.eq(2).any(axis=1)].eq(1).any(axis=1).all()

Output: False

To better understand how this works, let's define the following intermediates:

m1 = df.eq(2).any(axis=1)
m2 = df.shift()[m1].eq(1).any(axis=1)

Which gives us:

   A  B     m1     m2
0  1  3  False    NaN
1  2  3   True   True  # previous values are 1, 3 -> True
2  1  8  False    NaN
3  1  2   True   True  # previous values are 1, 8 -> True
4  3  5  False    NaN
5  2  7   True  False  # previous values are 3, 5 -> False
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1

If you find writing a loop easier to read rather than using some rather esoteric "one-liner" you could do this:

import pandas

df = pandas.read_csv("data.csv")
prev = next(gt := df.itertuples())

for current in gt:
    if (current.A == 2 and prev.A != 1) or (current.B == 2 and prev.B != 1):
        break
    prev = current
else:
    print("Yes")

Note: I have assumed that the range starting from 2 in the OP wasn't intended

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