Python: Iterate Over A Data Frame Column, Check For A Condition-value Stored In Array, And Get The Values To A List
Solution 1:
A quick way to do would be to leverage NumPy's broadcasting techniques as an extension of this answer from the same post linked, although an answer related to the use of DF.where
was actually asked.
Broadcasting eliminates the need to iterate through every element of the array and it's highly efficient at the same time.
The only addition to this post is the use of np.argmax
to grab the indices of the first True
instance along each column (traversing ↓ direction).
conditions = np.array([10, 15, 23])
tol = 0num_albums = df.Num_Albums.values
num_albums_cumsum = df.Num_Albums.cumsum().values
slices = np.argmax(np.isclose(num_albums_cumsum[:, None], conditions, atol=tol), axis=0)
Retrieved slices:
slices
Out[692]:
array([0, 2, 4], dtype=int64)
Corresponding array produced:
num_albums[slices]
Out[693]:
array([10, 4, 1], dtype=int64)
If you still prefer using DF.where
, here is another solution using list-comprehension
-
[df.where((df['cumsum'] >= cond - tol) & (df['cumsum'] <= cond + tol), -1)['Num_Albums']
.max() for cond in conditions]
Out[695]:
[10, 4, 1]
The conditions not fulfilling the given criteria would be replaced by -1. Doing this way preserves the dtype
at the end.
Solution 2:
well the output not always be 1 number right? in case the ouput is exact 1 number you can write this code
tol = 0
#condition
c = [5,15,25]
value = []
for i in c:
iflen(df.where((df['a'] >= i - tol) & (df['a'] <= i + tol)).dropna()['a']) > 0:
value = value + [df.where((df['a'] >= i - tol) & (df['a'] <= i + tol)).dropna()['a'].values[0]]
else:
value = value + [[]]print(value)
the output should be like
[1,2,3]
in case the output can be multiple number and want to be like this
[[1.0, 5.0], [12.0, 15.0], [25.0]]
you can use this code
tol = 5
c = [5,15,25]
value = []
for i in c:
getdatas = df.where((df['a'] >= i - tol) & (df['a'] <= i + tol)).dropna()['a'].values
value.append([x for x in getdatas])
print(value)
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