How to implement regex in if statement in dictionaries

Question

I have the following code snippet: I want to find the row, with the value is digit (they are integers all the time) on the column/key "Something" and delete the matched row. I have tried it with re.compile()

for row in list:
    if re.compile("\d") in row["Something"]:
        print(row)
        del (row)

How is the right implementation in this case?

do you need to filter out data containing digits, or data being all digits (like integers) ? — Jean-François Fabre, Nov 17 '17 at 18:38
Only remove the rows, with digits (they are all the time integers) — madik_atma, Nov 17 '17 at 18:49

Jean-François Fabre · Accepted Answer · 2017-11-17T19:26:27.267

1

don't try remove items from lists while iterating, it doesn't work that well: Remove items from a list while iterating. Anyway, del row has no effect since you're just deleting the name, not the item of the list.

I'm assuming that you want to remove rows if row["Something"] is an integer value.

I would rebuild the list using a list comprehension filtering out when the data is digits only (regex can be avoided here by using isdigit()):

[row for row in lst if not row["Something"].isdigit()]

That doesn't take care of negative numbers, mind. For that you would need a regex-type check like re.match("-?\d+",row["Something"])

edited Nov 17 '17 at 19:26

answered Nov 17 '17 at 18:32

Jean-François Fabre

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waste of time? talking with _me_ ? :) – Jean-François Fabre Nov 17 '17 at 18:42
that was a joke! – Jean-François Fabre Nov 17 '17 at 19:27
I got it, no worries :) – Jan Nov 17 '17 at 20:41

score 0 · Answer 2 · answered Nov 17 '17 at 18:35

Rather than deleting your row, you might want to match those which do not contain digits:

import re
rx = re.compile(r'^\D+$')

new_rows = (row for row in lst if rx.match(row))

This yields a generator which you could either iterate over or convert it to a list entirely.
Additionally, using list as a variable name is a no-no.

How to implement regex in if statement in dictionaries

2 Answers2