Question

在下面的代码中，每个x具有y的9个值。我想要new_gamesplyed = [x，y1，y2，y3，y4，y5，y6，y7，y8，y9]。但是在“差异”列表中，所有292个值都被立即存储。

[/home/VB] >cat common_functions.sh
#!/bin/sh

# the next line restarts using expect \
exec expect "$0" "$@"

proc wait_for_prompt {prompt {s 90}} {
    puts "wait_for_prompt calling ..."
    set timeout $s

    expect {
        $prompt {}
    }
}

proc handle_login { args } {
    puts "handle_login calling ..."

    expect {
        ":~#" {puts "handle_login accessed to remote"}
    }
}

[/home/VB] >cat test_expect_with_source.sh
#!/bin/sh

# the next line restarts using expect \
exec expect "$0" "$@"

source common_functions.sh

set remote_ip [lindex $argv 0]
set remote_remote_host [lindex $argv 1]
set timeout 5

puts "test_expect_with_source calling ..."

log_user 1
spawn ssh -o ConnectTimeout=3 root@$remote_ip

handle_login

puts "zone to access second remote"
after 1500
send "ssh $remote_remote_host\r"

wait_for_prompt ":*#"

...

[/home/VB] >./test_expect_with_source.sh 192.168.100.10 TESTING_MACHINE.TEST.HOST
test_expect_with_source calling ...
spawn ssh -o ConnectTimeout=3 root@192.168.100.10
handle_login calling ...
Last login: Wed Oct 23 07:42:25 2019 from 192.168.100.20
VB_machine:~# handle_login accessed to remote
zone to access second remote
wait_for_prompt calling ...    (confuse between this line)
ssh TESTING_MACHINE.TEST.HOST  (and this line)
...

Answer 1

您在整个代码段中都使用了相同的difference变量，因此也难怪它存储了append传递给它的所有值

如果我正确理解，您需要在主循环中的某些语句之间“重置”容器。因此，我建议每次在主循环语句开始时重新初始化存储桶集合。示例：

for x in range(len(df_gamesplayed)):      
    d_temp = []  # <---------  reset it here ----------
    for y in range (1, col_count):
        diff = abs(df_gamesplayed.iloc[x, y] - df_gamesplayed.iloc[x, y + 1])        
        d_temp.append(diff)
    new_gamesplayed.append([df_gamesplayed.iloc[x, 0], d_temp])

Python-将值追加到列表

1 个答案: