Python: re.findall result not appending to dictionary key

So I'm trying to go through a text file, grab the timings using re.findall, and then append the resulting list to the dictionary key. Here are the contents of the file:

Download Time: 1.234 sec

Commit Time: 4.321 sec

Reset Time: 6.96969 sec
***
Download Time: 8.313412 sec

Commit Time: 4.20420 sec

Reset Time: 9.699234 sec
***
Download Time: 5.678 sec

Commit Time: 2.3151 sec

Reset Time: 9.325346 sec
***

Now my issue is that I'm getting an empty dictionary result after the process is done. Not entirely sure why this is happening. Here's my code:

import re

timings = {
    "download": [],
    "commit": [],
    "reset": []
}   

# List of key phrases to look for in the text file
pattern = [r"Download Time: ([\d.]+) sec", 
           r"Commit Time: ([\d.]+) sec", 
           r"Reset Time: ([\d.]+) sec"]

# Loop through each key phrase and put timing data into proper list
# This is not well optimized lol (nested for loops)
for i in range(len(pattern)):
    with open(file, 'r') as f:
        # Open file contents and search for pattern
        content = f.read()
        matches = re.findall(pattern[i], content)

        # From collected matching phrases, append timing data into proper lists
        for key in timings:
            timings.update({key: matches})

The problem most likely lies in the last 2 lines of code. However, I'm not exactly sure what it would be since .update is a valid way to append a list into a key.

Here is another solution without changing your code a lot:

for key in pattern:
    with open(file, 'r') as f:
        # Open file contents and search for pattern
        content = f.read()
        matches = re.findall(key, content)

    # From collected matching phrases, append timing data into proper lists
    timings[key.split(" ")[0].lower()].extend(matches)

I changed the for loop in the beginning as you can see, so we don't use an index now.
About the last line, in order to take the name of the key, you split the regex pattern in sub-strings with " " (whitespace) as delimiter, you keep the first element, and then lower its letters.

For example:

The string: "Download Time: ([\d.]+) sec"

will break into: ['Download', 'Time:', '([\\d.]+)', 'sec'] whensplit(" ").

If we keep the first element (0-index) and lower it, we will have: download.
The same will happen for "commit" and "reset".

You can also use .partition(" ") instead of split(" ") which will break the whole string in 2 parts. The first is before the given delim. and the second is after it. In our case, delim. is whitespace.

Whatsoever, I would recommend you to learn to code solutions like the one @Barmar provided. My solution is just an alternative perspective if you would like to do smaller changes to the existing code.

Different approach which required named group patterns instead of usual group and finditer to iterate over the matched results.

Named groups are declared with the following syntax in a (?P<name>...).

The advantage is that the name of the group can be used as the key of the dictionary.

import re


text = # read from file


patterns = [r"Download Time: (?P<download>[\d.]+) sec", 
           r"Commit Time: (?P<commit>[\d.]+) sec", 
           r"Reset Time: (?P<reset>[\d.]+) sec"]


timings = {}
for p in patterns:
    for m in re.finditer(p, text):
        for key, v in m.groupdict().items():
            timings.setdefault(key, []).append(v)

print(timings)