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I have created the following pandas dataframe
import pandas as pd
import numpy as np
ds = {
'col1' :
[
['U', 'U', 'U', 'U', 'U', 1, 0, 0, 0, 'U','U', None],
[6, 5, 4, 3, 2],
[0, 0, 0, 'U', 'U'],
[0, 1, 'U', 'U', 'U'],
[0, 'U', 'U', 'U', None]
]
}
df = pd.DataFrame(data=ds)
The dataframe looks like this:
print(df)
col1
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None]
1 [6, 5, 4, 3, 2]
2 [0, 0, 0, U, U]
3 [0, 1, U, U, U]
4 [0, U, U, U, None]
For each row in col1, I need to check if every element equals to U in the list is followed (from left to right) by any value apart from U and None: in that case I'd create a new column (called iCount) with value of 1. Else 0.
In the example above, the resulting dataframe would look like this:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
Only in the first row the value U is followed by a value which is neither U nor None (it is 1)
I have tried this code:
col5 = np.array(df['col1'])
for i in range(len(df)):
iCount = 0
for j in range(len(col5[i])-1):
print(col5[i][j])
if((col5[i][j] == "U") & ((col5[i][j+1] != None) & (col5[i][j+1] != "U"))):
iCount += 1
else:
iCount = iCount
But I get this (wrong) dataframe:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 0
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
Can anyone help me please?
945
If you only want to test if there is at least one case in which a non-None follow a U, use itertools.pairwise and any:
from itertools import pairwise
def count_after_U(lst):
return int(any(a=='U' and b not in {'U', None} for a, b in pairwise(lst)))
df['iCount'] = list(map(count_after_U, df['col1']))
Output:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
5 [U, U, 4, U, U, 1, 0, U, U, None, 1, U, None] 1
6 [U, None, 1, U] 0
If you also want to check the other values until the next U, use a custom function:
def any_after_U(lst):
flag = False
for item in lst:
if item == 'U':
flag = True
else:
if flag and item is not None:
return 1
return 0
df['iCount'] = list(map(any_after_U, df['col1']))
Example:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
5 [U, U, 4, U, U, 1, 0, U, U, None, 1, U, None] 1
6 [U, None, 1, U] 1
UIIUC, use a custom python function:
from itertools import pairwise
def count_after_U(lst):
return sum(a=='U' and b not in {'U', None} for a,b in pairwise(lst))
df['iCount'] = list(map(count_after_U, df['col1']))
Or, to be more flexible with the conditions:
def count_after_U(lst):
flag = False
iCount = 0
for item in lst:
if item == 'U':
flag = True
else:
if flag and item is not None:
iCount += 1
flag = False
return iCount
df['iCount'] = list(map(count_after_U, df['col1']))
Output:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
More complex example:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
5 [U, U, 4, U, U, 1, 0, U, U, None, 1, U, None] 2
U:Just indent the flag reset in the previous approach to only reset it if a value was not yet found:
def count_after_U(lst):
flag = False
iCount = 0
for item in lst:
if item == 'U':
flag = True
else:
if flag and item is not None:
iCount += 1
flag = False
return iCount
df['iCount'] = list(map(count_after_U, df['col1']))
Example:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
5 [U, U, 4, U, U, 1, 0, U, U, None, 1, U, None] 3
Try this:
def calcUs(lst):
cnt = 0
for x, y in zip(lst, lst[1:]):
if (x == 'U' and y != 'U' and y != None):
cnt += 1
return cnt
df['iCount'] = df['col1'].apply(lambda x: calcUs(x))
df
Output:
col1 iCount
0 [U, U, U, U, U, 1, 0, 0, 0, U, U, None] 1
1 [6, 5, 4, 3, 2] 0
2 [0, 0, 0, U, U] 0
3 [0, 1, U, U, U] 0
4 [0, U, U, U, None] 0
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